Science Ncert Exemplar 2019 Solutions for Class 9 Science Chapter 4 - Structure Of Atom

Answer:

The atomic number of magnesium is 12 which means a magnesium atom has 12 electrons. So, the electronic configuration can be written as  2, 8, 2.

Hence, the correct answer is option (b).

Answer:

During Rutherford’s α-particle scattering experiment, some of the alpha particles are deflected through small and large angles. This observation shows that there is a 'centre of positive charge' in the atom which repels the positively charged alpha particles and deflects them from their original path. This centre of positive charge in the atom is known as the nucleus. Therefore, Rutherford’s α-particle scattering experiment led to the discovery of the atomic nucleus.

Hence, the correct answer is option (c).

Answer:

The correct representation of an element is given by ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$
where, 
A is mass number of the element
X is symbol of the element
Z is the atomic number of the element

For the given element X,
Atomic number = Number of electrons = 15 
Mass number = 15 + 16 = 31

Thus, the representation of element X is ${}_{15}{}^{31}\mathrm{X}$. 

Hence, the correct answer is option (a).

Answer:

Dalton’s atomic theory successfully explained the law of conservation of mass, law of constant composition, and law of multiple proportions.

Hence, the correct answer is option (d).

Answer:

Rutherford's model of atom explained the presence of a nucleus in the centre and electrons around the nucleus revolving in round orbitals like the solar system. The positively charged alpha particles were deflected by the nucleus. This shows the nucleus is positively charged.

Hence, the correct answer is option (a).

Answer:

The formula for the atomic number and mass number can be written as:
Atomic number = number of protons = number of electrons
Mass number = Number of protons + Number of neutrons

Hence, the correct answer is option (d).

Answer:

Thomson's model could be compared with a raisin pudding model according to which the mass of the atom is uniformly distributed over the atom in the form of positive charge and electrons are uniformly distributed over the atom. Electrons do not attract each other.
Hence, the correct answer is option (a).

Answer:

The following conclusions were drawn by Rutherford from $\mathrm{\alpha }$-particle scattering experiment:
(i) Most of the space inside the atom is empty.
(ii) All the positive charge and mass of the atom were concentrated in a very small volume within the atom.
(iii) The positive charge of the atom occupies very little space.

Hence, the correct answer is option (b).

Answer:

Mass number of the atom = 27
Number of neutrons = 14
Number of protons = 27 − 14 = 13
Number of electrons in the atom = 13
Number of electrons in ion with 3 positive charges = 13 − 3 = 10

Hence, the correct answer is option (b).

Answer:

A magnesium atom can be represented as :  ${}_{12}{}^{24}\mathrm{Mg}$
For magnesium atom,
Atomic number = Number of  protons = Number of electrons = 12
Number of neutrons = Mass number − Number of protons = 24 − 12 = 12
If magnesium loses 2 electrons, it forms Mg²⁺ ion.
Number of electrons in Mg²⁺= 10

Hence, the correct answer is option (d).

Answer:

The isotope of an element contains the same number of electrons and protons but a different mass number. As the number of electrons and number of protons are equal in an atom, a different mass number means a different number of neutrons. 

Hence, the correct answer is option (c).

Answer:

The metals which have 1 valence electron and non-metals which have 7 valence electrons can exhibit a valency of 1. Therefore, both metal and non-metals can show a valency of 1.

Hence, the correct answer is option (c).

Answer:

J.J. Thomson was the first one to propose a model for the structure of an atom.

Hence, the correct answer is option (d).

Answer:

For the atom
Number of neutrons = 4
Number of protons = 3
Therefore, the number of electrons = 3
Hence, the distribution of electrons = 2,1
Since the atom can lose 1 electron to achieve the inert gas electron configuration, therefore, the valency of the given atom is 1.

Hence, the correct answer is option (c).

Answer:

Aluminium has 13 electrons. Its electron distribution can be written as 2, 8, 3.

Hence, the correct answer is option (a).

Answer:

According to Bohr's model of an atom, the first shell can have a maximum of 2 electrons and the second shell can have a maximum of 8 electrons. Fig. (ii) contains 4 electrons in K shell, and fig. (iv) contains 9 electrons in L shell, which are not in accordance with Bohr's model. 

Hence, the correct answer is option (c).

Answer:

An atom has an equal number of electrons and protons since it is electrically neutral.

Hence, the correct answer is option (a).

Answer:

The correct chronological order of the improvements in atomic models is Thomson's atomic model, followed by Rutherford's atomic model which is followed by Bohr's atomic model.

Hence, the correct answer is option (c).

Answer:

Yes, it is possible. Hydrogen has one electron, one proton and no neutron.

Answer:

Two observations which support the fact that atoms are divisible are:
(i) Ionic compounds are formed because of the formation of ions that involves the transfer of electrons.
(ii) The presence of isotopes for the same element is possible due to the difference in the number of neutrons.
The above observations show that an atom is formed by different particles such as electrons, protons, and neutrons, i.e. atom is divisible.

Answer:

No,  ³⁵Cl and ³⁷Cl have the same valencies. They are isotopes of chlorine and have the same number of protons and electrons. They have same electron distribution, therefore, the same valency.

Answer:

Gold is a heavy metal with a high mass number and hence the fast-moving alpha particles will not be able to push the atoms and will scatter after colliding with the gold atoms. Also, gold is a highly malleable element can be beaten into very thin sheets.

Answer:

(a) The electronic configuration of the atom is 2,8,8. Its outermost shell has a complete octet. Hence, its valency is zero.
(b) The electronic configuration of the atom is 2,7. Its outermost shell has an incomplete octet. It needs to gain one electron to complete its octet. Hence, its valency is 1.

Answer:

An element X is a metal because one electron is present in the outermost shell. When this valence electron is removed from the outermost shell, a cation (positive ion) will be formed with a charge of +1.
$\mathrm{X} - 1{\mathrm{e}}^{-} \to {\mathrm{X}}^{+}$
 

Answer:

The atomic number of chlorine is 17. The electronic distribution of chlorine is 2, 8, 7. Hence it has 2 electrons in its innermost shell, 8 electrons in its second shell, and 7 electrons in the outermost shell respectively. Therefore, the L shell of chlorine contains 8 electrons.

Answer:

An atom having 6 electrons in its outermost shell will require two electrons to complete the octet.
$\mathrm{X} + 2{\mathrm{e}}^{-} \to {\mathrm{X}}^{2-}$
The ion formed will have a negative charge of −2.

Answer:

Answer:

The given statement is incorrect. The number of protons is never greater than the number of neutrons. The number of neutrons can be equal to or greater than the number of protons because a mass number is equal to double the atomic number or greater than double the atomic number. Also, number of neutrons can be greater than the number of electrons because the number of electrons and protons is always equal in a neutral atom.

Answer:

For the given element X,
Number of protons = Number of electrons = 15
Number of neutrons = Mass number − Number of protons = 31 − 15 = 16
Therefore, the given element X has 16 neutrons.

Answer:

(a) - (iii)
(b) - (iv)
(c) - (i)
(d) - (ii)
(e) - (vi)
(f) - (vii)
(g) - (v)

Answer:

Two or more elements having the same mass number but a different atomic number are known as isobars. Therefore, ${}_{20}{}^{40}\mathrm{Ca}$ and ${}_{18}{}^{40}\mathrm{Ar}$ are pair of isobars.

Answer:

Answer:

Helium has only one shell (K shell) and the maximum number of electrons within the K shell can be 2. Thus, its shell is already complete called duplet. It can neither lose electrons nor gain electrons. Hence, its valency is zero. It is called noble gas or inert gas.

Answer:

(a) nucleus
(b) atomic number, mass number
(c) 0 and 1
(d) silicon- 2,8,4 ; sulphur- 2,8,6

Answer:

The element X has atomic number 2. Thus, an atom of X contains two electrons in it. These two electrons fill the K shell completely. Thus, the valency of X is zero.

Answer:

Helium has two electrons which completely fill the K shell. Neon has ten electrons which completely fill K and L shells. Similarly, Argon has 18 electrons which completely K, L and M shell. 
The valence shells of these elements are completely filled, thus, they all have stable electronic configuration, and hence, their valency is zero.

Answer:

(i) Volume of the sphere = $\frac{4}{3}{\mathrm{\pi r}}^3$
Let R be the radius of the atom and r be that of the nucleus.
 
$\frac{\mathrm{R}}{\mathrm{r}}={10}^5$ (given)

⇒ R = 10⁵r 

 $\begin{array}{rcl}\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{atom} & =& \frac{4}{3}{\mathrm{\pi R}}^3=\frac{4}{3}\mathrm{\pi }{\left({10}^5\mathrm{r}\right)}^3\\ & =& \frac{4}{3}{\mathrm{\pi r}}^3\times {10}^{15}\end{array}$

$\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{nucleus} = \frac{4}{3}{\mathrm{\pi r}}^3$
The ratio of the size of the atom to that of the nucleus = $\frac{{\displaystyle \frac{4}{3}}\times {10}^{15}\times {\mathrm{\pi r}}^3}{{\displaystyle \frac{4}{3}}{\mathrm{\pi r}}^3}={10}^{15}$

(ii) If the atom is represented by the planet earth R_e = 6.4 ×10⁶ m

Radius of nucleus = r_n = $\frac{{\mathrm{R}}_{\mathrm{e}}}{{10}^5}$

 $\begin{array}{rcl}{\mathrm{r}}_{\mathrm{n}} & =& \frac{6.4\times {10}^6 \mathrm{m}}{{10}^5}\\ & =& 64 \mathrm{m}\\ & & \end{array}$

Answer:

Rutherford concluded from α-ray scattering experiment that
1. Most of the space in the atom is empty because most of the α-particles passed through the gold foil without getting deflected.
2. A small fraction of α-particles were deflected indicating that the positive charge of the atom occupies very little space. 
3. A very small fraction of α-particles were deflected by 180º, indicating that all the positive charge and mass of the atom were concentrated in a very small volume within the atom.

Answer:

Thomson's model of the atom proposed the resin pudding structure in which electrons are embedded like resins in a positively charged sphere of pudding or cake which neutralise the positive charge so that, the atom as a whole is electrically neutral.
Rutherford's model could be compared with the solar system. According to his model, the positive charge is concentrated in a very small space in the centre which was called the nucleus. Electrons revolve around the nucleus in well-defined orbits.

Answer:

Drawbacks of Rutherford’s model of an atom:
1. A major drawback (or defect) of Rutherford's model of the atom is that it does not explain the stability of the atom.
Any particle in a circular motion would undergo acceleration and would radiate energy. Thus revolving electrons would lose energy and finally fall into the nucleus. Thus, the atom should be unstable and should not exist in a stable form. Rutherford's model, however, could not explain the stability of an atom.
2. The model did not give any information about the arrangement of electrons around the nucleus.

Answer:

According to Bohr's model of the structure of the atom
1. In an atom, the electrons revolve rapidly around the nucleus in fixed circular paths called orbits or shells.
2. Only certain special orbits are allowed inside the atom.
3. While revolving in discrete orbits the electrons do not radiate energy. They absorb energy to go to a higher level and emit energy to go to the lower level. These orbits are called energy levels which are represented as K, L, M, N or the numbers, n =1,2,3,4.
 

Answer:

Atomic number (Z) of sodium is 11.
Electronic configuration = 2, 8, 1

Sodium ion (Na⁺) is formed by the loss of one electron from the valence shell.
Electronic configuration = 2, 8 


 

Answer:

$$\begin{gathered} \% \mathrm{of} \mathrm{\alpha }-\mathrm{particles} \mathrm{deflected} \mathrm{more} \mathrm{than} 50° = 1\% \\ \% \mathrm{of} \mathrm{\alpha }-\mathrm{particles} \mathrm{deflected} \mathrm{less} \mathrm{than} 50° = (100 - 1)\% = 99\% \\ \mathrm{Number} \mathrm{of} \mathrm{\alpha }-\mathrm{particles} \mathrm{bombarded} = 1 \mathrm{mole} = 6.022\times {10}^{23} \mathrm{particles} \\ \mathrm{Number} \mathrm{of} \mathrm{\alpha }-\mathrm{particles} \mathrm{deflected} \mathrm{at} \mathrm{an} \mathrm{angle} \mathrm{less} \mathrm{than} 50° = \frac{99}{100}\times 6.022\times {10}^{23} \\ = \frac{596.178 }{100}\times {10}^{23} = 5.96 \times {10}^{23} \end{gathered}$$