Among the following the odd pair is
A. 188O, 3717Cl
B. 4018Ar, 147N
C. 3014Si, 3115P
D. 5424Cr, 3919K
It can be seen that the third pair has atoms with same number of neutrons i.e. the two atoms are isotones. But, none other pair has isotenes.
Change in the number of neutrons in an atom changes it to
A. an ion.
B. an isotope.
C. an isobar.
D. another element.
Isotope is defined as the elements having similar mass number. And we know that
Mass Number = Total no. Of Protons+ Total no. of Neutrons
Thus, if mass no. is same then the no. of neutrons must be same.
The term nucleons refer to
A. Protons and electrons
B. only Neutrons
C. electrons and neutrons
D. Protons and neutrons
Nucleons are referred to the protons and the neutrons.
The number of protons, neutrons and electrons present respectively in 8035Br
A. 80, 80, 35
B. 35, 55,80
C. 35, 35, 80
D. 35, 45, 35
Atomic number = Total no. of electrons = 35
Now, No. of electrons = No. of Protons = 35
And, Mass Number = Protons + Neutrons = 80
No. of neutrons = Mass no. – Protons
= 80 – 35 = 45
The correct electronic configuration of potassium is
A. 2,8,9
B. 2,8,1
C. 2,8,8,1
D. 2,8,8,3
No. of electrons in Potassium = 19
No. of electrons in K-shell = 2
L-shell = 8
M-shell = 8
N-shell = 1
Calculate the volume of oxygen required for the complete combustion of 20 cm3 of methane[CH4(g) + 2O2 → CO2(g) +2H2O(g)].
Thus, the Volume of oxygen used is 40 cm3
A metal combines with oxygen to form two oxides having the following composition
i) 0.398 gram of metal oxide I contains 0.318 gram of metal
ii) 0.716 gram of metal oxide II contains 0.636 gram of metal. So that the above data agrees with the law of multiple proportions.
i) 0.398g metal oxide contains 0.318g of metal.
Therefore, amount of oxygen combining with
0.318g metal oxide I = 0.398-0.318 = 0.080g
ii) 0.716g metal oxide contains 0.636g of metal.
Therefore, amount of oxygen combining with
0.318g metal oxide I = 0.716-0.636 = 0.080g
Thus, for fixed mass of oxygen (0.080g), the metal forma oxides in the ratio = 0.318:0.636 = 1:2
Since, it is a simple ratio therefore the above data agrees with the law of multiple proportion.
Calculate the mass of a proton, given its charge = + 1.60 × 10−19 C
charge / mass = 9.58 × 107 C kg−1
Given:
Charge of proton = + 1.60 × 10−19 C
And, = 9.58 × 107 C kg−1
Thus, the mass of proton is
What conclusions were made from the observations of Gold foil experiment?
From theGold foil experiment, Rutherford and his team observed that:
● Most of the fast moving α-particles passed straight through the gold foil.
● Some α particles were deflected by small angles and a few by large angles.
● Surprisingly very few α particles completely rebounded.
Thus the conclusions made were:
● Atom has a very small nucleus at the centre.
● There is large empty space around the nucleus.
● Entire mass of an atom is concentrated in a very small positively charged region which is called the nucleus.
● Electrons are distributed in the vacant space around the nucleus.
● The electrons move in circular paths around the nucleus.
Explain the postulates of Bohr’s atomic model.
The main postulates of Bohr’s atomic model are:
● In atoms, electrons revolve around the nucleus in certain special or permissible orbits known as orbits or shells or energy levels.
● While revolving in these orbits the electrons do not radiate energy.
● The circular orbits are numbered as 1, 2, 3, 4, … or named as K, L, M, N, shells. These numbers are referred to as principal quantum numbers (n).
● K shell (n = 1) is closest to the nucleus and is associated with lowest energy. L, M, N, …. etc are the next higher energy levels. As the distance from the nucleus increases the energy of the shells also increases.
● The energy of each orbit or shell is a fixed quantity and the energy is quantized.
● As the distance from the nucleus increases, the size of the orbits also increases.
● Maximum number of electrons that can be accommodated in an energy level is given by 2n2 where n is the principal quantum number of the orbit.
● When an electron absorbs energy, it jumps from lower energy level to higher energy level.
● When an electron returns from higher energy level to lower energy level, it gives off energy.
State the Gay Lussac’s law of combining volumes, explain with an illustration.
Gay Lussac’s law of combining volumes:
This law states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.
The law explains experimental facts about how gaseous atoms combine.
Example:
For the reaction: N2(g) + 3H2(g)→ 2NH3(g)
1 vol. 3 vols. 2 vols.
1 volume of nitrogen combines with 3 volumes of hydrogen to form 2 volumes of ammonia.
Thus, the ratio is 1:3:2.
This satisfies the Gay Lussac’s law of combining volumes.
The particles represented above are
1. a) and d) Electrons
2. d) Neutrons
3. d) Protons
From the structures given below, Tabulate the following:
1. Valence electron
2. Valency
3. Atomic Number
4. Mass number
5. Electronic configuration
Figure1:
1. Valence electron: 4
2. Valency: 4
3. Atomic Number: 6
4. Mass number: 12
5. Electronic configuration: 2, 4
Figure2:
1. Valence electron: 2
2. Valency: 2
3. Atomic Number: 12
4. Mass number: 24
5. Electronic configuration: 2, 8, 2
Figure3:
1. Valence electron: 7
2. Valency: 1
3. Atomic Number: 17
4. Mass number: 35
5. Electronic configuration: 2, 8, 7
Figure4:
1. Valence electron: 2
2. Valency: 2
3. Atomic Number: 20
4. Mass number: 40
5. Electronic configuration: 2, 8, 8, 2
Figure5:
1. Valence electron: 1
2. Valency: 1
3. Atomic Number: 19
4. Mass number: 39
5. Electronic configuration: 2, 8, 8, 1
Figure6:
1. Valence electron: 3
2. Valency: 3
3. Atomic Number: 13
4. Mass number: 27
5. Electronic configuration: 2, 8, 3
The correct numbers of protons and neutrons present in 2311Na+ are
c) 11, 12
(Since, the protons and neutrons remains same)
In an atom, electrons revolve around the nucleus in fixed orbits.
True
Yes, the electrons revolve around the nucleus in fixed orbits.
Isotopes of an element have the different atomic numbers.
False
Isotopes of elements have different mass numbers.
Electrons have negligible mass and charge.
False
Electrons have negligible mass but is has a charge of 1.602 x 10-19 Coulombs.
Smaller the size of the orbit, lower is the energy of the orbit.
True
Since, as we go higher the energy starts increasing. Thus higher the orbit higher will be the energy and lower the orbit lower will be the energy.
The maximum number of electron in L Shell is 10.
False
The no. of electrons in L shell is 8.
Since, No. of electrons = 2n2 = 2(2)2 = 2x4 = 8
Calcium and Argon are examples of a pair of __________.
Isobars
Total Number of electrons that can be accommodated in an orbit is given by __________.
2n2
__________ isotope is used in the treatment of goitre.
iodine-131
The number of neutrons present in 73LI is ___________.
4
The valency of Argon is ___________.
0
Since, argon is a noble gas thus its valency is zero.
Match the following:
Match the following:
Complete the following table
*Deuterium is an isotope of Hydrogen
Arrange the following in the increasing order of atomic number
Calcium, Silicon, Boron, Magnesium, Oxygen, Helium, Neon, Sulphur, Fluorine and Sodium
Rearranging in the increasing order of atomic number:
1. Helium
2. Boron
3. Oxygen
4. Fluorine
5. Neon
6. Sodium
7. Magnesium
8. Silicon
9. Sulphur
10. Calcium
Cross word puzzle
Clues:
Down:
1. Helium Nuclei (Particle)
2. Positive Charge mass at the core of the atom
3. An atom whose valency is zero
4. One or two electrons in the outermost shell of atoms of elements are called as ______________ electrons.
5. 146C is used for carbon dating
6. Discovery of neutron
Across:
1. Electrons present in the outermost shell
2. This pair of atoms 4020Ca, 4018Al are ____
3. An atom that does not have neutron
4. Scattering of α particles in the gold foil experiment
Down
1. Alpha
2. Hydrogen
3. Argon
4. Valence
5. Carbon
6. Chadwick
Across:
1. Valence
2. Lone Pair
3. Hydrogen
4. Rutherford
Copy the following and write the names of the laws and their simple definitions in the space provided.
● Box-1: Law Of Multiple Proportions:
This law states that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.
● Box-2: Law of Conservation of Mass:
This law states that the mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations.
According to the law of conservation of mass, the total mass of the reaction in a chemical reaction must equal the mass of the reactants.
● Box-3: Law of constant Proportions:
The law of constant composition says that, in any particular chemical compound, all samples of that compound will be made up of the same elements in the same proportion or ratio.
● Box-4: Law of Reciprocal proportions:
This law states that if two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine.
● Box-5: Gay Lussac’s Law of Gaseous volumes:
This law states that the ratio between the volumes of the reactant gases and the gaseous products can be expressed in simple whole numbers. 2 molecules of hydrogen + 1 molecule of oxygen = 2 molecules of water.
Name an element which has the same number of electrons in its first and second shell.
Berilium (atomic number: 4)
Write the electronic configuration of K+ and Cl–
Electronic configuration of K+:
K-shell: 2
L-shell: 8
M-shell: 8
(Since one electron is lost thus the electronic configuration will be like this)
Electronic configuration of Cl-:
K-shell: 2
L-shell: 8
M-shell: 8
(Since one electron is added thus the electronic configuration will be like this)
Compare the charge and mass of protons and electrons.
Charge:
The electronic charge of the proton and the electron both are same. Only there is a change on sign.
Since proton has a positive charge thus its charge is +1.6 × 10-19 C
Whereas electron has negative charge so the charge of electron is -1.6 × 10-19 C.
Mass:
The proton is much heavier in weight whereas electrons have negligible weight.
The weight of proton is 1.67 × 10-24 g whereas the mass of the electron is 9.31 × 10-28 g.
For an atom ‘X’, K, L and M shells are completely filled. How many electrons will be present in it?
We know that maximum no of electrons in:
K-shell = 2
L-shell = 8
M-shell = 18
Therefore; total no. of electrons = 2+8+18 = 28.
Ca2+ has completely filled outer shell. Justify your answer.
Ca2+ means that calcium has lost 2 electrons.
Now; atomic number of Ca = 20
Since, it has lost 2 electrons, thus the no of electrons left are = 18
Thus, atomic configuration of Ca2+ = 2, 8, 8
This shows that the Ca2+ has a completely filled outer shell.
State the law of multiple proportions.
Law of Multiple Proportions:
This law states that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.
List the uses of isotopes?
Uranium235: This isotope of uranium is used as fuel in nuclear reactor.
Cobalt60: This isotope of cobalt is used in treatment of cancer.
Iodine131: This isotope of iodine is used in treatment of goitre.
Carbon14: This isotope of carbon is used in Carbon dating (that is to find the age of plants and animals).
What is isotone? Give an example
Isotone is defined as two or more atoms having the same number of neutrons.
Example: chlorine-37 and potassium-39 are isotones, because the nucleus of this species of chlorine consists of 17 protons and 20 neutrons, whereas the nucleus of this species of potassium contains 19 protons and 20 neutrons.
Draw the structure of oxygen and sulphur atoms.
Structure of oxygen atom:
Structure of sulphur atom:
Calculate the number of neutrons, protons and electrons (i) atomic number 3 and mass number 7 (ii) atomic number 92 and mass number238.
i) Given,
Atomic Number = 3
Mass Number = 7
Therefore,
No. of electrons = atomic number = no. of protons = 3
And,
Mass Number = no. of protons+ no. of neutrons = 7
Thus, no. of neutrons = 4
ii) Given,
Atomic Number = 92
Mass Number = 238
Therefore,
No. of electrons = atomic number = no. of protons = 92
And,
Mass Number = no. of protons+ no. of neutrons = 238
Thus, no. of neutrons = 238-92 = 146