Chemical Bonding And Molecular Structure

Isostructural species are those which have the same shape and hybridization.

NF₃ is pyramidal whereas BF₃ is planar triangular.
BF^–₄ and NH⁺₄ ions both are tetrahedral.
BCl₃ is a triangular planar whereas BrCl₃ is pyramidal.
NH₃ is pyramidal whereas NO₃ is a triangular planar.

Thus, option (ii) is correct.

H₂O will have the highest dipole moment due to the maximum difference in electronegativity of H and O atoms.

Thus, option (iii) is correct.

The number of orbitals involved in hybridization can be determined by the application of formula:

H = number of orbitals involved in hybridization
V= valence electrons of the central atom
M= number of monovalent atoms linked with the central atom
C = charge on the cation
A = charge on the anion

Thus, for NO₂⁺ H is 2 i.e. sp, for NO₃^- H is 3 i.e. sp² and for NH₄⁺ H is 4 i.e. sp³.

The strength of H-bonding depends on the electronegativity of the atom which follows the order: F > O > N.
Thus, the strength of bonding will be in the following order:
H……. F > H…….O > H…….. N
The H₂O molecule is linked to 4 other H₂O molecules through H-bonds whereas each HF molecule is linked only to two other HF molecules.

Thus, the H₂O molecules bonding will be stronger than HF bonding.

Hence, correct decreasing order of the boiling points is H₂O > HF > NH₃

The formal charge on an atom can be calculated from the following formula:

Thus, for oxygen atom in PO₄^3- ion the formal charge is -1.

Formal charge on oxygen atom = 6-(6+) = -1

In the given structure one can see there are 4 bonds and 0 electron pairs on the nitrogen molecule.

The shape of all the compounds can be calculated from hybridization, which can be determined by the application of formula:

H = number of orbitals involved in hybridization
V= valence electrons of a central atom
M= number of monovalent atoms linked with the central atom
C = charge on the cation
A = charge on the anion

Thus, the shape of all molecules are

BH^-₄ is tetrahedral, NH^-₂ is V-shape, CO₃^2- is triangular planar and H₃ O⁺ is pyramidal.

Each double bond has one pie bond. The single bond is sigma bond.

So, we can see in the above diagram that Napthalene molecule has 5 double bond and 19 sigma bonds.

Thus, Option (iii) is the correct answer.

The O₂^2- has no unpaired electron. The O₂ molecules have six electrons in the valence shell it accepts two-electron to complete its octet.

Thus, The O₂^2- has no unpaired electron.

All the other molecules have an incomplete octet. Thus, they contain unpaired electrons.

C₂H₄ has one double bond and four single bonds. The bond length of double bond (C = C) is smaller than single bond (C – H).

The other compounds have single bonds only that have same bond length.

The strength of H-bonding depends on the electronegativity of the atoms.
The H₂O molecule is linked to 4 other H₂O molecules through H-bonds and all other compounds don’t form hydrogen bonding.

Thus, the H₂O molecules bonding will be stronger than other compounds.

The outermost orbital is also known as valence orbital and the electrons present in that orbital participate in the bond formation.

For the given electronic-configuration the d and s orbitals are the outermost orbitals.

Thus, the 4 electrons involved in chemical bond formation will be

3d², 4s².

The hybridization sp² has orbital s donating 33.33% percent to the structure and 2 orbitals of p, donating rest of the percentage to the molecule i.e. 66.66%.

Thus, the shape of the molecule is trigonal planar. Where it is angled at 120⁰ degrees to each other.

Hybridization is the process of orbital mixing.

The last orbital of the element A has completed octet (i.e. 8 electrons in the last energy shell (orbital s and p)) so it is in a stable state.

The element C requires 1 electron to complete its octet so it will share the outer electrons will another C atom.

Thus, the compound formed after the sharing of electrons will be C₂.

The atom B requires 3 electrons to complete its octet whereas C atom requires only 1 electron.

Thus, 3 atoms of C will combine with a B atom to get a stable electronic configuration.

The compound will be BC₃.

Both the B and C atoms are non-metallic, hence they will share the pair of an electron to complete the octet.

Thus, the bond between them will be covalent in nature.

The Molecular Orbital diagram is different for molecules with 14 or less electrons than the one used for molecules with 15 or more electrons.

For N₂ the orbitals in increasing energy are:

(π2p_y ) < (σ2pz ) < (π* 2px) ≈ (π* 2p_y)

All the statements (i),(ii) and (iii) are correct

Statement (iv) is incorrect because according to the Molecular Orbital Theory the increasing order of orbitals in term of energy is:

(π2p_y ) < (σ2pz ) < (π* 2px) ≈ (π* 2p_y)

MOT of O₂ is given below:

MOT of O₂⁺ is given below:

Bond Order of O₂ is 2.0

Bond Order of O₂⁺ is 2.5

Bond Order of O₂^- is 1.5

Thus, correct order is O₂^- < O₂ < O₂⁺.

Group 17 or VIIA is called the most electronegative group because of their small atomic radius due to which high nuclear attraction from the relatively more no. of protons than the no. of orbits/shells.

All the Group 17 elements have outermost ns² np⁵ configuration.

The electronic configuration represents:
2s²2p⁵= fluorine = most electronegative element
3s²3p⁵ = chlorine
4s²4p⁵ = bromine
5s²5p⁵ = iodine

Thus, option (i) is correct.

The ionization energy is the amount of the energy required to remove an electron from the valence orbital of the atom.

The ionization enthalpy depends on the atomic size, smaller the atomic radius the greater is the ionization enthalpy.

In option (i) the nuclear charge is on the last electron is very less so it will have less ionization energy compare to the option (ii).

(ii) and (iv) have exactly half-filled p-orbitals but (ii) is smaller in size.

Thus, [Ne]3s²3p³ has the highest ionization enthalpy.

The isoelectronic species have a similar bond order.

Both CN^- and NO⁺ have 14 electrons.

Both CN^- and NO⁺ have identical bond order i.e. 3.

(MOT diagram of CN^-)

So, from the above diagram in CN^-,

Electrons in bonding molecular orbitals = 8

Electrons in Anti Bonding Molecular orbitals = 2

Bond Order = .

= (8-2)

= 3

So, from the above MOT diagram of NO⁺,

Electrons in bonding molecular orbitals = 10(2e^- in 1s)

Electrons in Anti Bonding Molecular orbitals = 4(2e^- in 1*s)

Bond Order = .

= (10-4)

= 3.

The structure of BeCl₂ and CS₂ is linear in shape.

Both Be and C atom has sp hybridization.

(Linear structure of CS₂ )

(Linear structure of BeCl₂)

NO₂ is bent due to the interelectronic repulsion.

(Bent structure of NO₂)

Isoelectronic species are the atoms of different elements with the same number of electrons.

Both (i), (ii) have 14 electrons and CO also has 14 electrons (6 electrons of C and 8 electrons of Oxygen).

SnCl₂ has 84 electrons (50 electrons of Sn and 34 electrons of Cl₂ ) and NO₂ has 23 electrons (7 electrons of N and 16 electrons of O₂).

The structure of CO₂ is linear, CCl₄ is tetrahedral in structure, O₃ is bent shaped and NO₂^- is linear in shape.

(Linear structure of NO₂^-)

(Linear structure of CO₂)

Thus, CO₂ and NO₂^- have the same structure.

The hybridization of Carbon atom is sp² in carbonate ion.

Carbonate ion is present in the form of a resonating structure. These structures are equivalenting to nature. Resonance all 3 C-O bonds get a double character in one of the resonating structures.

Thus, all the bonds are equivalent and have equal length.

Diamagnetic species have no unpaired electrons in the molecular orbital. These species repel the magnetic field.

The N₂ molecule has no unpaired electron in the valence shell/molecular orbital. Thus, it is diamagnetic in nature.

N₂^2- and O₂ have unpaired electron so they are paramagnetic in nature.

The bond order of N₂=

The bond Order of N₂^-=

The bond Order of F₂⁺=

The bond Order of O₂==1.5

Thus, options (iii) and (iv) have the same bond order.

The ionic compounds are good conductors of electricity in the molten state. Thus, the statement (i) is incorrect.

In the canonical structure, there is a difference in the arrangements of electrons. Thus, the statement (ii) is incorrect.

The statement (iii) and statement (iv) is correct.

In H₂S, the Sulphur atom is surrounded by the four electron pairs (two bond pairs and two lone pairs).

These four electron pairs adopt tetrahedral geometry.

The repulsion between the lone pair electrons brings distortion in the shape of the H₂S.

Thus, H₂Sis not linear in shape.

In PCl₃, the phosphorous atom has one lone pair of electron and three bond pairs. The hybridization is sp³ i.e tetrahedral shape.

The presence of the lone pair brings distortion in the tetrahedral shape.

Thus, it is pyramidal in shape.

The Molecular Orbital configuration of O₂⁺ and O^-₂ is given below:

O₂⁺ (15): σ1s² σ ^*1s² σ 2s² σ ^*2s² σ 2p_z² π2p_x² = π 2p_y²π ^*2p_x¹

O₂^- (17): σ1s² σ ^*1s² σ 2s² σ ^*2s² σ 2p_z² π2p_x² = π 2p_y²π ^*2p_x²= π^*2p_y¹

Bond order for O₂⁺

Bond order for O^-₂

According to Molecular Orbital Theory, the greater the bond order greater is the bond energy.

Thus, O₂⁺ is more stable than O^-₂.

Both of the ions have unpaired electrons. Thus, they will be paramagnetic in nature.

In BrF₅ the central atom is bromine which has the hybridization sp³d².

Br atom has 7 valence electrons out of which 5 are used to make pair with the F atoms and two are used to make lone pair of electrons.

The lone pair and bond pair repel each other. Thus, the shape is square pyramidal.

The white balls are F atoms and the pink ball is Br atom.

(a) Compound (I)

Compound (II)

(b) The compound (II) has a higher melting point because of the intermolecular bonding, a large number of molecules that will get attached to each other.

(c) The compound (II) will be more soluble in water because it will form hydrogen bonding with the water molecules easily. On another hand, the compound (I) has intramolecular hydrogen bonding so it will be difficult for it to combine with H₂O molecules.

in the figure (i) the area of the contact of ++ overlap is equal to the area of the +- overlap. The so-net overlap is zero.

In figure (ii) there is no overlap of the orbitals due to the different symmetry.

In PCl₅, P has 5 valence electrons in orbitals. To make 5 bonds with 5 Cl atoms, it will share one of its electrons from 3s to 3d orbital, therefore the hybridization will be sp³d.

And with sp³d hybridization, the geometry will be trigonal bipyramidal.

IF₅, the Iodine atom has 7 valence electrons in molecular orbitals.

Thus, Iodine will form 5 bonds with 5 Cl atoms using 5 electrons from its molecular orbital, two electrons will form one lone pair on Iodine atom, which gives the square pyramidal geometry to IF₅ molecule.

Dimethyl ether will have a greater bond angle. There will be more repulsion between bond pairs of CH₃ groups attached in ether than between bond pairs of hydrogen atoms attached to oxygen in the water.

The carbon of CH₃ in ether is attached to three hydrogen atoms through σ bonds and electron pairs of these bonds add to the electronic charge density on the carbon atom. Hence, the repulsion between two —CH₃ groups will be more than that between two hydrogen atoms.

The formal charge is calculated:

The formal charge on the oxygen with single bond =

The formal charge on the oxygen with double bond

The formal charge on nitrogen=

The formal charge on oxygen 1 and 4 =

The formal charge on oxygen 2 and 3 =

The formal charge on hydrogen 1 and 2 =

The formal charge on sulfur =

The general sequence of the energy level of molecular orbitals are:

The molecular orbitals in the sequence of their energy levels for the given molecules are given below:

The bond order is given by the formulae:

Bond order for N₂=

Bond order for N₂⁺=

Bond order for N₂^-

Bond order for N₂²⁺

According to Molecular Orbital Theory, the greater the bond order greater is the bond energy.

Thus, the order of stability is:

N₂> N₂^- > N₂⁺> N₂²⁺

N₂^- is more stable than N₂⁺ because it has more bonding electrons i.e. electrons in the bonding molecular orbital.

The molecules or ions with unpaired electrons are paramagnetic else diamagnetic in nature.

Thus, N₂ and N₂²⁺ are diamagnetic and N₂⁺ and N₂^- are paramagnetic in nature.

(i) N₂ is having 14 electrons when it donates one electron, these electrons are removed from the Bonding molecular orbital.

Thus,

The bond order reduces.

(ii) O₂ has 16 electrons, 8 electrons in the molecular orbitals and 4 in the antibonding molecular orbitals.

It donates one electron from the anti-molecular bonding orbital.

Thus,

The bond order increases.

(i) A covalent bond is formed by the overlapping of atomic orbitals. The direction of overlapping gives the direction of the bond.

In an ionic bond, the electrostatic field of an ion is non-directional. Each positive ion is surrounded by a number of anions in any direction depending upon its size and vice-versa.
That’s why covalent bonds are directional bonds while ionic bonds are non-directional.

(ii) In a water molecule, the oxygen atom is sp³ hybridized and has two lone pairs of electrons.

The shape acquired by 4 molecular orbital is tetrahedral geometry with two hydrogen atoms occupying the corner and the other two orbitals by lone pair of electrons.

The bond angle is reduced to 104.5° from 109.5° due to greater repulsive forces between lone pairs. Thus, a molecule thus acquires a V-shape or bent structure (angular structure).

In carbon dioxide molecules, the carbon molecule is sp hybridized.

The two sp orbitals are oriented in opposite direction to each other. Thus, the shape is linear.

(iii) In the ethyne molecule, both the carbon atoms are sp hybridized. The two sp hybrid orbitals of both the carbon atoms are oriented in the opposite direction forming an angle of 180°.

Thus, the molecule is linear in shape.

The definition of an ionic bond is when a positively charged ion forms a bond with a negatively charged ions and one atom transfers electrons to another. An example of an ionic bond is the chemical compound Sodium Chloride (NaCl).

The difference between an ionic and a covalent bond are:

(a) An ionic bond essentially donates an electron to the other atom participating in the bond, while electrons in a covalent bond are shared equally between the atoms.

(b)The only pure covalent bonds occur between identical atoms. Usually, there is some polarity (polar covalent bond) in which the electrons are shared, but spend more time with one atom than the other.

(c) Ionic bonds form between a metal and a non-metal. Covalent bonds form between two non-metals.

To arrange the molecules in the increasing order of ionic character, we have to look at the electronegativity difference.

The ionic character is greater in the molecules that are having the greatest electronegativity difference because the electron pair shifts toward a more electronegative atom increasing the ionic character.

The electronegativity difference order in the molecules is as follows:

C-H<N-H<O-H< F-H

Thus, the ionic character order will be:

C-H<N-H<O-H< F-H

Carbonate ion is present in the form of a resonating hybrid structure. These structures are equivalent in nature. Resonance all 3 C-O bonds get a double character in one of the resonating structures.

Thus, all the bonds are equivalent and have equal length hence carbonate ion cannot be represented by a single Lewis structure.

The hybridization of Carbon 1 is sp, carbon 2 is sp, carbon 3 sp², carbon 4 is sp³ and carbon 5 is sp².

The triple bond has 2 pie bonds and one sigma bond.

Each double bond has one sigma and one pie bond.

Every single bond is a sigma bond.

Thus, the total number of sigma bonds is 11 and pie bonds are 4.

BeCl₂ has a linear structure because Be has no lone pairs. Thus, the electrons on the chlorides will try to stay far apart from each other, since their corresponding electrons repel each other (while experiencing no deflection from electrons on a central atom).

HOCl is also non-linear in structure. It is unstable in form and, the Cl atom has sp³ hybridized, occurs as Cl-O^- ion which has a linear structure.

H₂O has a V-shaped structure. It results from the sp³ -hybridization of O -atom.

Cl₂O has a V-shaped structure. The oxygen atom undergoes sp³ hybridization.

(i); XH₄, H₃Y, and HZ

Hydrogen has only one electron in its outermost shell it shares one electron to form a covalent bond or accepts or donates one electron to form an ionic bond.

For XH₄, the X element requires 4 electrons to complete its octet so 4 hydrogen atoms will form a covalent bond with X similarly for element Y 3 atoms of Hydrogen will form a covalent bond to form compound HY₃.

For HZ, the bond will be ionic in nature hydrogen will donate one electron to form cation and Z will accept one electron to form anion and both of them will form an ionic bond.

(ii); The compound HZ has a linear shape and the difference in the electronegativity of Hydrogen and element Z is maximum.

Z has maximum electronegativity due to the small atomic radius and greater nuclear charge.

The resonating structures of the Ozone and Nitrate molecules are:

In compound BCl₃, Boron has sp²-hybridisation and the shape is Triangular Planar.

In methane CH₄, Carbon has sp³ -hybridization and shape are Tetrahedral.

In carbon dioxide CO₂, carbon has sp-hybridisation and shape is Linear.

In ammonia NH₃, nitrogen has sp³-hybridisation and shape is Pyramidal.

Carbonate ion is present in the form of a resonating structure. These structures are equivalenting to nature. Resonance all 3 C-O bonds get a double character in one of the resonating structures.

Thus, all the bonds are equivalent and have equal length.

Average bond Enthalpy is because all the similar bonds in a molecule do not have the same bond enthalpies.

Therefore, in polyatomic molecules term mean or average bond enthalpy is used. It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken.

For water, the two OH bonds have different bond enthalpies.

This is the average bond enthalpy of O-H bond in water.

Thus, there is a difference in bond enthalpy of the O—H bond in ethanol (C₂H₅OH) and water.

(i)(c);

The hybridisation of the S is sp³d.

No. of Hybrid Orbitals = (V+M-C+A)

= (6+4)

= 5(As it has 5 hybridised orbitals).

So, it’s hybridisation is sp³d.

(ii)→ (a);

The hybridization of the I is sp³d².

Iodine has valency = 7.

No. of Hybrid Orbitals = (7+5)

= 6

Thus, It’s hybridisation is sp³d².

(e);

As, there is 1 (+) charge.

No. of Hybrid Orbitals = (5-1)

=2.

The hybridization of the N is sp

(iv) (d);

As, there is 1 cation.

No. of hybrid orbitals = (5+4-1)

= 4.

The hybridization of the N is sp³

(i) (e);

The shape of H₃O⁺ is Tetrahedral. The hybridization is sp³.

(ii) (a);

As Ethylene has double bond in its structure.

So, Ethylene is linear in shape. The hybridization is sp.

(iii) (b);

ClO^-₂ is angular in structure due to lone pair and bond pair repulsion.

(iv) (c);

NH⁺₄ is tetrahedral in shape and the hybridization of N is sp³.

Thus, on calculating we can get the bond order:

(i) (c);

So, Bond Order of NO = (15-10)

= 2.5

(ii) (d);

So, Bond Order of CO = (14-8)

= 3.

(iii) (a);

So, Bond Order of O₂^- = (10-7)

= 1.5

(iv) (b);

So, Bond Order of O₂ = (16-12)

= 2.

(i) (d);

the HF compound has hydrogen bonding due to a difference in the electronegativity of hydrogen and fluorine.

(ii) (e);

Ozone molecules have resonating structures.

(iii) (b);

LiF is an ionic compound Li donates one electron and fluorine accepts one to complete its octet.

(iv) (a);

Carbon is a covalent solid due to its sharing of electron and catenation property it forms agiant molecule with 3-d network.

(i) (c);

the sp³ hybridized molecules have a tetrahedral shape.

(ii) (a);

the sp² hybridized molecules have a trigonal shape.

(iii) (b);

the sp hybridized molecules have a linear shape.

(i);

The sodium chloride is a stable compound. It has an ionic bond. The sodium metal donates one electron and chlorine accepts one to complete its octet.

Thus, both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

(i);

The central atom of both NH₃ and H₂O molecules are sp³ hybridized, yet H–N–H bond angle is greater than that of H–O–H. This is because the nitrogen atom has one lone pair and the oxygen atom has two lone pairs. The lone pair and bond pair repulsion is greater in the ammonia as compared to water.

Thus, both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

(iv);

Both the assertion and the reason is false. There is average bond enthalpy for the O-H bonds of water.

(i) Dipole moment plays a very important role in understanding the nature of the chemical bonds.

A few applications are given below:

(1) Distinction between, polar and non-polar molecules. The measurement of the dipole moment can help us to distinguish between polar and non-polar molecules. Non-polar molecules have zero dipole moment while polar molecules have some value of dipole moment.

(2) Degree of polarity in a molecule. Dipole moment measurement also gives an idea about the degree of polarity, especially in a diatomic molecule. The greater the dipole moment, the greater is the polarity in such a molecule.

(3) The shape of molecules. In the case of molecules containing more than two atoms, the dipole moment not only depends upon the individual dipole moments of the bonds but also on the arrangement of bonds. Thus, the dipole moment is used to find the shapes of molecules.

(4) Distinguish between cis- and trans- isomers. Dipole moment measurements help to distinguish between cis- and trans- isomers because ds-isomer has usually higher dipole moment than trans isomer.

(5) Distinguish between ortho, meta and para isomers. Dipole moment measurements help to distinguish between o-, m- and p-isomers because the dipole moment of p-isomer is zero and that of o-isomers is more than that of m-isomer.

(ii) The diagrammatical representation of bond moments and the resultant dipole moment in CO₂, NF_3, and CHCl₃ are:

NF₃

CHCl₃

The general sequence of the energy level of molecular orbitals for nitrogen is:

The molecular orbitals in the sequence of their energy levels for the given molecules are given below:

Since the bond order is 3, N₂ has a triple bond.

F₂ = σ 1s², σ ^* 1s², σ 2s², σ^* 2s², σ 2p_x², π 2p_x² = π2p_y²

The molecular orbital of Fluorine is given

The bond order of F₂ =

Thus, when bond order is 1, the number of bonds is also 1.

Ne₂ = σ 1s², σ ^* 1s², σ 2s², σ^* 2s², σ 2p_x², π 2p_x² = π2p_y², π^*2p_x² = π^*2p_y²

The bond order of Ne₂==0

Thus, Ne₂ has no bond.

The valence bond theory is based on the knowledge of atomic orbitals and electronic configurations of elements, overlap criteria of atomic orbitals and stability of the molecule.

The main points of the valence bond theory are:

(i) Atoms do not lose their identity even after the formation of the molecule.

(ii) The bond is formed due to the interaction of only the valence electrons as the two atoms come close to each other. The inner electrons do not participate in bond formation.

(iii) During the formation of a bond, only the valence electrons from each bonded atom lose their identity. The other electrons remain unaffected.

(iv) The stability of the bond is accounted for by the fact that the formation of the bond is accompanied by release of energy. The molecule has minimum energy at a certain distance between the atoms known as inter-nuclear distance. Larger the decrease in energy, stronger will be the bond formed.

Consider two hydrogen atoms A and B approaching each other having nuclei H and H and the corresponding electrons e^- and e^- respectively.

When atoms come closer to form molecules new forces begin to operate.

(a) The force of attraction between the nucleus of the atom and electron of another atom.

(b) The force of repulsion between two nuclei of the atom and electron of two atoms.

When two hydrogen atoms are at a farther distance, there is no force operating between them, when they start coming closer to each other, the force of attraction comes into play and their potential energy starts decreasing. As they come closer to each other potential goes on decreasing, but a point is reached when potential energy acquires minimum value.

(a) This distance corresponding to this minimum energy value is called the distance of maximum possible approach, i.e. the point which corresponds to minimum energy and maximum stability.

(b) If atoms come further closer than this distance of maximum possible approach, then potential energy starts increasing and the force of repulsion comes into play and molecules start becoming unstable.

The hybridization of the P in PCl₅ is sp³d and the hybridization of the S in SF₆ is sp³d².

PCl₅ has trigonal bipyramidal geometry. In this case, the axial bonds are slightly longer than the equatorial bonds. This is because the axial bonds experience greater repulsion from other bonds than the equatorial bonds.

SF₆ has an octahedral structure. In this case, all the bonds have the same length since all bonds experience similar repulsion from other bonds.

(i) The hybridization is defined as the process of the intermixing of the orbitals of the slightly different or same energy level to form new orbitals of similar shape and energy level. These orbitals are called hybrid orbitals.

The valence shell of an atom is always hybridized and the merging orbitals are of nearly similar energy or same energy.

These orbitals do not form pie bonds.

(a) sp hybridization: This type of hybridization is found in the carbon compounds having triple bond i.e. .

(b) sp² hybridization: This type of hybridization is found in the carbon compounds having a double bond i.e. C=C.

(c) sp³ hybridization: this type of hybridization is found in the carbon compounds having single bonds i.e. C-C.

(ii)

(a)

(b)

(c)

(d)

(e)

(i)

Two molecular orbitals in the dioxygen have unpaired electrons.

Thus, statement (ii) is incorrect.

The total number of bonding molecular orbitals is the same as the total number of antibonding orbitals in dioxygen.

Thus, the statement (iii) is incorrect.

The number of filled bonding orbitals will not be always the same as a number of filled antibonding orbitals.

Thus, the statement (iv) is incorrect.

Thus, statement (i) is correct i.e. in the formation of dioxygen from oxygen atoms 10 molecular orbitals will be formed.

(iv);

The σ*1s, σ*2p_z, and π2p_x have only one nodal plane while π*2p_y orbital has 2 nodal planes.

The nodal plane divides the molecular orbitals.

(ii);

The Isoelectronic species have a similar bond order.

Both O⁺₂ and N₂^- has 15 electrons.

O₂ has 16 electrons and N₂ has 14 electrons.

O₂^- has 17 electrons and N⁺₂ has 13 electrons.

O^-₂ has 17 electrons and N^-₂ has 15 electrons.

(iii);

In the atoms having electron less than or equal to the 14 electrons, the σ2p_z molecular orbital is filled after π2p_x and π2p_y molecular orbitals.

Only nitrogen in the above options has 14 electrons other options have more than 14 electrons.

Thus, in nitrogen, the σ2p_z molecular orbital is filled after π2p_x and π2p_y molecular orbitals.