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NCERT Exemplar · Class 9 Science Structure of the Atom

43 questions · 43 still being checked

Short Answer Questions 29–35 (part 4 of 5)

  1. Exercise 29

    Calculate the number of neutrons present in the nucleus of an element X which is represented as 31\displaystyle 31 15\displaystyle 15 X

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    NCERT’s answer
    Mass number = No. of protons + No. of neutrons = $\displaystyle 31$ ∴ Number of neutrons = $\displaystyle 31$- number of protons = $\displaystyle 31$-$\displaystyle 15$ = $\displaystyle 16$
    \[A = 31, \quad Z = 15 \] \[N = A - Z = 31 - 15 = 16 \] Answer: \(\displaystyle 16\) neutrons.

  2. Exercise 30

    Match the names of the Scientists given in column A with their contributions
    towards the understanding of the atomic structure as given in column B
    (A)
    (B)(a)
    Ernest Rutherford
    (i)
    Indivisibility of atoms
    (b)
    J.J.Thomson
    (ii)
    Stationary orbits
    (c)
    Dalton
    (iii)
    Concept of nucleus
    (d)
    Neils Bohr
    (iv)
    Discovery of electrons
    (e)
    James Chadwick
    (v)
    Atomic number
    (f)
    E. Goldstein
    (vi)
    Neutron
    (g)
    Mosley
    (vii)
    Canal rays

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    NCERT’s answer
    (a)
    (iii)(b)
    (iv) (c) (i) (d) (ii) (e) (vi) (f) (vii) (g) (v)
    Rutherford's \(\displaystyle \alpha\)-particle scattering gave the nucleus. J.J. Thomson found the electron. Dalton held atoms indivisible. Bohr fixed stationary orbits. Chadwick found the neutron. Goldstein observed canal rays. Mosley defined atomic number.Answer: (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii), (e)-(vi), (f)-(vii), (g)-(v).

  3. Exercise 31

    The atomic number of calcium and argon are 20\displaystyle 20 and 18\displaystyle 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements?

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    NCERT’s answer
    Isobars
    \[Z_{\text{Ca}} = 20, \quad Z_{\text{Ar}} = 18 \] \[A_{\text{Ca}} = A_{\text{Ar}} = 40 \] Same mass number, different atomic number.Answer: Isobars.

  4. Exercise 32

    Complete the Table 4.1\displaystyle 4.1 on the basis of information available in the symbols given below:
    (a)
    1735Cl\displaystyle ^{35}_{17}\text{Cl}
    (b)
    612C\displaystyle ^{12}_{6}\text{C}
    (c)
    3581Br\displaystyle ^{81}_{35}\text{Br}
    Table 4.1\displaystyle 4.1 -- Element, np\displaystyle n_p, nn\displaystyle n_n.

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    NCERT’s answer
    Element \(\displaystyle n_{p}\) \(\displaystyle n_{n}\) Cl C Br
    For \(\displaystyle ^{A}_{Z}\text{X}\), \(\displaystyle n_p = Z\) and \(\displaystyle n_n = A - Z\).
    (a)
    \(\displaystyle ^{35}_{17}\text{Cl}\): \[n_p = 17\] \[n_n = 35 - 17 = 18\]
    (b)
    \(\displaystyle ^{12}_{6}\text{C}\): \[n_p = 6\] \[n_n = 12 - 6 = 6\]
    (c)
    \(\displaystyle ^{81}_{35}\text{Br}\): \[n_p = 35\] \[n_n = 81 - 35 = 46\]
    Answer: Cl: \(\displaystyle n_p = 17\), \(\displaystyle n_n = 18\); C: \(\displaystyle n_p = 6\), \(\displaystyle n_n = 6\); Br: \(\displaystyle n_p = 35\), \(\displaystyle n_n = 46\).

  5. Exercise 33

    Helium atom has 2\displaystyle 2 electrons in its valence shell but its valency is not 2\displaystyle 2, Explain.

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    NCERT’s answer
    Helium atom has $\displaystyle 2$ electrons in its outermost shell and its duplet is complete. Hence the valency is zero.
    \[\text{He}:\ Z = 2 \] \[2n^2 = 2(1)^2 = 2 \] The K shell holds at most $\displaystyle 2$ electrons; helium's $\displaystyle 2$ valence electrons already complete this duplet, so helium need not lose, gain or share any electron.Answer: Valency \(\displaystyle =0\) — the complete duplet gives no tendency to combine.

  6. Exercise 34

    Fill in the blanks in the following statements
    (a)
    Rutherford’s α-particle scattering experiment led to the discovery of
    the ———
    (b)
    Isotopes have same ———but different———.
    (c)
    Neon and chlorine have atomic numbers 10\displaystyle 10 and 17\displaystyle 17 respectively. Their
    valencies will be———and———respectively.
    (d)
    The electronic configuration of silicon is ———and that of sulphur

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    NCERT’s answer
    (a)
    atomic nucleus (b) atomic number, mass number (c) $\displaystyle 0$ and 1. (d) Silicon—$\displaystyle 2$, $\displaystyle 8$, $\displaystyle 4$ Sulphur— $\displaystyle 2$, $\displaystyle 8$, $\displaystyle 6$
    (a)
    \(\displaystyle \alpha\)-particle scattering \(\displaystyle \Rightarrow\) nucleus.
    (b)
    same atomic number, different mass number.
    \[\text{Ne }(Z=10):\ K,L = 2,8 \Rightarrow \text{valency } 0 \]
    \[\text{Cl }(Z=17):\ K,L,M = 2,8,7 \Rightarrow \text{valency } 1 \]
    \[\text{Si }(Z=14):\ K,L,M = 2,8,4 \]
    \[\text{S }(Z=16):\ K,L,M = 2,8,6 \]
    Answer: (a) nucleus; (b) atomic number, mass number; (c) $\displaystyle 0$ and $\displaystyle 1$; (d) $\displaystyle 2,8,4$ and $\displaystyle 2,8$,6.

  7. Exercise 35

    An element X has a mass number 4\displaystyle 4 and atomic number 2. Write the valency of this element?

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    NCERT’s answer
    Valency is zero as K shell is completely filled.
    \[A = 4,\ Z = 2 \Rightarrow \text{X is } \text{He} \] \[\text{electrons} = Z = 2, \quad 2n^2 = 2(1)^2 = 2 \] The single K shell is full — a complete duplet, so X need not lose, gain or share any electron.Answer: Valency \(\displaystyle =0\).