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NCERT Exemplar · Class 9 Science Structure of the Atom

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Short Answer Questions 19–28 (part 3 of 5)

  1. Exercise 19

    Is it possible for the atom of an element to have one electron, one proton and no neutron. If so, name the element.

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    Yes, it is true for hydrogen atom which is represented as $\displaystyle 1$ 1H
    Yes — the ordinary isotope of hydrogen has exactly this composition. \[A = p + n = 1 + 0 = 1, \qquad Z = p = 1 \] \[\Rightarrow {}^{1}_{1}\text{H} \] Answer: Yes, hydrogen (protium), \(\displaystyle ^{1}_{1}\text{H}\).

  2. Exercise 20

    Write any two observations which support the fact that atoms are divisible.

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    Hint— Discovery of electrons and protons
    Cathode rays (fast negative particles) stream from the cathode toward the anode in a discharge tube — atoms give up negatively charged electrons. Perforated-cathode tubes show canal rays travelling the opposite way, carrying positive charge — atoms also give up positively charged protons. Both particles come from inside the atom, so the atom cannot be an indivisible sphere. Answer: discovery of the electron (cathode rays) and the proton (canal rays).

  3. Exercise 21

    Will 35Cl and 37Cl have different valencies? Justify your answer.

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    Hint— No, 35Cl and 37Cl are isotopes of an element.
    \[Z(^{35}\text{Cl}) = Z(^{37}\text{Cl}) = 17 \] \[\text{Electron distribution (both): } 2,\,8,\,7 \] Isotopes differ only in neutron number; the proton count, and so the electron arrangement, is identical. \[\text{Valence electrons} = 7 \ \Rightarrow\ \text{valency} = 8-7 = 1 \] Valency is fixed by the outer-shell electrons, which the extra neutrons never touch. Answer: no — both \(\displaystyle ^{35}\text{Cl}\) and \(\displaystyle ^{37}\text{Cl}\) have valency \(\displaystyle 1\).

  4. Exercise 22

    Why did Rutherford select a gold foil in his α-ray scattering experiment?

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    Hint— gold has high malleability
    Gold is exceptionally malleable, so it can be beaten into a foil about $\displaystyle 1000$ atoms thick — thinner than any other easily available metal, keeping the number of atomic layers an \(\displaystyle \alpha\)-particle must cross to a minimum. Gold is also chemically unreactive, so the foil's surface stays clean and undistorted. Answer: Malleable \(\displaystyle \Rightarrow\) thinnest possible foil, minimal multiple scattering; inert \(\displaystyle \Rightarrow\) clean, stable target.

  5. Exercise 23

    Find out the valency of the atoms represented by the Fig. 4.3\displaystyle 4.3 (a) and (b).
    (a)
    (b)
    NCERT_Question_Class9_Science_Exemplar_Ch4_Q23

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    (a)
    $\displaystyle 0$ (b) $\displaystyle 1$
    (a) Valency = 0. The shells hold \(\displaystyle 2, 8, 8\) electrons — the outermost shell is complete, so the atom neither loses nor gains an electron.(b) Valency = 1. The shells hold \(\displaystyle 2, 7\) electrons — one short of the octet, so one electron is gained: \(\displaystyle 8-7=1\).Answer: (a) $\displaystyle 0$; (b) 1.

  6. Exercise 24

    One electron is present in the outer most shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outer most shell?

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    + $\displaystyle 1$
    \[\text{Neutral atom: } p = e \] \[\text{Removing the sole outer electron: } e \to e-1 \] \[\text{Net charge} = (+p) + \big(-(e-1)\big) = +1 \] The ion is a cation carrying one unit of positive charge. Answer: \(\displaystyle \text{X}^{+}\), charge \(\displaystyle +1\).

  7. Exercise 25

    Write down the electron distribution of chlorine atom. How many electrons are there in the L shell? (Atomic number of chlorine is 17\displaystyle 17).

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    NCERT’s answer
    $\displaystyle 2$, $\displaystyle 8$, 7. The L shell has eight electrons
    \[Z(\text{Cl}) = 17 \] \[K = 2,\quad L = 8,\quad M = 17-2-8 = 7 \] NCERT_Solution_Class9_Science_Exemplar_Ch4_Q25 Each shell fills to its stable count before the next begins, so the electron distribution is \(\displaystyle 2,8,7\). Answer: \(\displaystyle 2,8,7\); the L shell holds \(\displaystyle 8\) electrons.

  8. Exercise 26

    In the atom of an element X, 6\displaystyle 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed?

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    -$\displaystyle 2$
    Element X's outermost shell has $\displaystyle 6$ electrons, $\displaystyle 2$ short of the stable $\displaystyle 8$-electron octet. To reach noble-gas configuration it gains $\displaystyle 2$ electrons.\[\text{charge} = -(\text{electrons gained}) = -2 \]Answer: \(\displaystyle -2\); the ion formed is \(\displaystyle \text{X}^{2-}\).

  9. Exercise 27

    What information do you get from the Fig. 4.4\displaystyle 4.4 about the atomic number, mass number and valency of atoms X, Y and Z? Give your answer in a tabular form. NCERT_Question_Class9_Science_Exemplar_Ch4_Q27

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    Atomic No. Mass No. Valency X Y Z $\displaystyle 3,5$
    \[A = n_p + n_n \] \[A_X = 5+6=11,\quad A_Y = 8+10=18,\quad A_Z = 15+16=31 \]Valency is the outermost-shell electron count when \(\displaystyle \le 4\); when \(\displaystyle >4\) it is usually \(\displaystyle 8-\)(that count), but a $\displaystyle 5$-electron outer shell (Z) also shows the count itself, so Z's valency is both $\displaystyle 3$ and 5.
    AtomAtomic numberMass numberElectron configurationValency
    X$\displaystyle 5$$\displaystyle 11$$\displaystyle 2$, $\displaystyle 3$$\displaystyle 3$
    Y$\displaystyle 8$$\displaystyle 18$$\displaystyle 2$, $\displaystyle 6$$\displaystyle 2$
    Z$\displaystyle 15$$\displaystyle 31$$\displaystyle 2$, $\displaystyle 8$, $\displaystyle 5$$\displaystyle 3$, $\displaystyle 5$
    Answer: X: $\displaystyle 5$, $\displaystyle 11$, $\displaystyle 3$; Y: $\displaystyle 8$, $\displaystyle 18$, $\displaystyle 2$; Z: $\displaystyle 15$, $\displaystyle 31$, $\displaystyle 3$ or 5.

  10. Exercise 28

    In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement? Justify your answer.

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    Hint— No, the statement is incorrect. In an atom the number of protons and electrons is always equal.
    \[\text{Neutral atom, net charge} = (+p) + (-e) = 0 \ \Rightarrow\ p = e \] Protons must equal electrons in every neutral atom — never one greater than the other. The claim orders \(\displaystyle p > n > e\), which needs \(\displaystyle p>e\); that step is impossible, so the statement is wrong regardless of how \(\displaystyle n\) compares to either. Answer: no — \(\displaystyle p=e\) always, so \(\displaystyle p\) can never exceed \(\displaystyle e\).