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NCERT Exemplar · Class 10 Science Periodic Classification of Elements

52 questions · 52 still being checked

Short Answer Questions 27–36 (part 4 of 6)

  1. Exercise 27

    The three elements A, B and C with similar properties have atomic masses X, Y and Z respectively. The mass of Y is approximately equal to the average mass of X and Z. What is such an arrangement of elements called as? Give one example of such a set of elements.

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    NCERT’s answer
    The arrangement of these elements is known as Dobereiner triad. Example, Lithium, Sodium and Potassium
    Döbereiner's triad \[Y \approx \dfrac{X+Z}{2} \] Example: \(\displaystyle \mathrm{Li,\ Na,\ K}\), atomic masses $\displaystyle 7$, $\displaystyle 23$, $\displaystyle 39$: \[\dfrac{7+39}{2} = 23 = \text{atomic mass of } \mathrm{Na} \] Answer: Döbereiner's triad; e.g. \(\displaystyle \mathrm{Li,\ Na,\ K}\).

  2. Exercise 28

    Elements have been arranged in the following sequence on the basis of their increasing atomic masses.
    F, Na, Mg, Al, Si, P, S, Cl, Ar, K
    (a)
    Pick two sets of elements which have similar properties.
    (b)
    The given sequence represents which law of classification of elements?

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    NCERT’s answer
    (a)(i)
    F and Cl
    (ii)
    Na and K.
    (b)
    Newland's law of octaves
    (a)
    Similar properties:
    \[\mathrm{F,\ Cl}\ (\text{halogens}) \qquad \mathrm{Na,\ K}\ (\text{alkali metals}) \]
    (b)
    Atomic masses increase along the sequence and similar properties recur: Newlands' Law of Octaves.
    Answer: (a) \(\displaystyle \mathrm{F,\ Cl}\); \(\displaystyle \mathrm{Na,\ K}\). (b) Newlands' Law of Octaves.

  3. Exercise 29

    Can the following groups of elements be classified as Dobereiner's triad ?
    (a)
    Na, Si, Cl
    (b)
    Be, Mg, Ca
    Atomic mass of Be 9\displaystyle 9; Na 23\displaystyle 23; Mg 24\displaystyle 24; Si 28\displaystyle 28; Cl 35\displaystyle 35; Ca 40\displaystyle 40
    Explain by giving reason.

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    NCERT’s answer
    (a)
    No, because all these elements do not have similar properties although the atomic mass of silicon is average of atomic masses of sodium (Na) and chlorine (Cl).
    (b)
    Yes, because they have similar properties and the mass of magnesium (Mg) is roughly the average of the atomic mass of Be and Ca.
    (a)
    \(\displaystyle \mathrm{Na,\ Si,\ Cl}\) are a metal, a metalloid and a non-metal, so their properties differ. The mean fits:
    \[\dfrac{23+35}{2} = 29 \approx 28\ (\mathrm{Si}) \]
    but similarity of properties, not the mean alone, is required. Not a triad.
    (b)
    \(\displaystyle \mathrm{Be,\ Mg,\ Ca}\) are all alkaline-earth metals, and
    \[\dfrac{9+40}{2} = 24.5 \approx 24\ (\mathrm{Mg}) \]
    A triad.
    Answer: (a) No, the properties differ; (b) Yes.

  4. Exercise 30

    In Mendeléev 's Periodic Table the elements were arranged in the increasing order of their atomic masses. However, cobalt with atomic mass of 58.93\displaystyle 58.93 amu was placed before nickel having an atomic mass of 58.71\displaystyle 58.71 amu. Give reason for the same.

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    NCERT’s answer
    Hint— Elements with similar properties can be grouped together.
    Similarity of properties took precedence over mass order: \(\displaystyle \mathrm{Co}\) was placed first so that elements with similar properties fall in the same group. \[\text{Atomic mass: } \mathrm{Co}\ 58.93\ \mathrm{amu} > \mathrm{Ni}\ 58.71\ \mathrm{amu} \] \[\text{Atomic number: } \mathrm{Co}\ 27 < \mathrm{Ni}\ 28 \] Answer: Properties were given priority over atomic mass; by atomic number \(\displaystyle \mathrm{Co}\) ($\displaystyle 27$) correctly precedes \(\displaystyle \mathrm{Ni}\) ($\displaystyle 28$).

  5. Exercise 31

    "Hydrogen occupies a unique position in Modern Periodic Table". Justify the statement.

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    NCERT’s answer
    Hint— Hydrogen resembles alkali metals as well as halogens
    Like the Group $\displaystyle 1$ alkali metals, \(\displaystyle \mathrm{H}\) has one valence electron and forms a unipositive ion: \[\mathrm{H \rightarrow H^{+} + e^{-}} \] Like the Group $\displaystyle 17$ halogens, it exists as a diatomic molecule \(\displaystyle \mathrm{H_2}\) and gains one electron to form the hydride ion: \[\mathrm{H + e^{-} \rightarrow H^{-}} \] It sits in Group $\displaystyle 1$ yet resembles Group $\displaystyle 17$, so its position is unique. Answer: \(\displaystyle \mathrm{H}\) resembles both the alkali metals (\(\displaystyle \mathrm{H^+}\)) and the halogens (\(\displaystyle \mathrm{H_2}\), \(\displaystyle \mathrm{H^-}\)).

  6. Exercise 32

    Write the formulae of chlorides of Eka-silicon and Eka-aluminium, the elements predicted by Mendeléev.

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    NCERT’s answer
    \(\displaystyle \mathrm{GeCl}_4, \mathrm{GaCl}_3\)
    Eka-silicon (Es) lies below \(\displaystyle \mathrm{Si}\) in the same group, valency $\displaystyle 4$: \[\mathrm{EsCl_4} \quad (\text{now known as } \mathrm{GeCl_4}) \] Eka-aluminium (Ea) lies below \(\displaystyle \mathrm{Al}\) in the same group, valency $\displaystyle 3$: \[\mathrm{EaCl_3} \quad (\text{now known as } \mathrm{GaCl_3}) \] Answer: \(\displaystyle \mathrm{EsCl_4\ (GeCl_4)}\) and \(\displaystyle \mathrm{EaCl_3\ (GaCl_3)}\).

  7. Exercise 33

    Three elements A, B and C have 3\displaystyle 3, 4\displaystyle 4 and 2\displaystyle 2 electrons respectively in their outermost shell. Give the group number to which they belong in the Modern Periodic Table. Also, give their valencies.

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    NCERT’s answer
    ElementGroup No.Valency
    AGroup-$\displaystyle 13$$\displaystyle 3$
    BGroup-$\displaystyle 14$$\displaystyle 4$
    CGroup-$\displaystyle 2$$\displaystyle 2$
    Group-number rule for a main-group element: with $\displaystyle 1$–$\displaystyle 2$ valence electrons the group number equals that count; with $\displaystyle 3$–$\displaystyle 8$, add ten. \[A:\ n=3 \Rightarrow \text{Group } 13,\ \text{valency } 3 \] \[B:\ n=4 \Rightarrow \text{Group } 14,\ \text{valency } 4 \] \[C:\ n=2 \Rightarrow \text{Group } 2,\ \text{valency } 2 \] Answer: A: Group $\displaystyle 13$, valency $\displaystyle 3$; B: Group $\displaystyle 14$, valency $\displaystyle 4$; C: Group $\displaystyle 2$, valency 2.

  8. Exercise 34

    If an element X is placed in group 14\displaystyle 14, what will be the formula and the nature of bonding of its chloride?

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    NCERT’s answer
    \(\displaystyle \mathrm{XCl}_4\); Covalent bonding
    Group $\displaystyle 14$ gives \(\displaystyle \mathrm{X}\) four valence electrons, so valency $\displaystyle 4$: \[\mathrm{X + 2Cl_2 \rightarrow XCl_4} \] With four electrons to share, \(\displaystyle \mathrm{X}\) bonds covalently, as carbon and silicon do in \(\displaystyle \mathrm{CCl_4}\) and \(\displaystyle \mathrm{SiCl_4}\). Answer: \(\displaystyle \mathrm{XCl_4}\); covalent.

  9. Exercise 35

    Compare the radii of two species X and Y. Give reasons for your answer.
    (a)
    X has 12\displaystyle 12 protons and 12\displaystyle 12 electrons
    (b)
    Y has 12\displaystyle 12 protons and 10\displaystyle 10 electrons

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    NCERT’s answer
    Hint— Radii of Y is less than X because Y is cation of X
    \[X=\mathrm{Mg}:\ 2,8,2 \qquad Y=\mathrm{Mg^{2+}}:\ 2,8 \] Y has lost the outer shell ($\displaystyle 3$ shells \(\displaystyle \to\) $\displaystyle 2$) while the nucleus still holds $\displaystyle 12$ protons. \[12\,p^+ \text{ on } 12\,e^- \to 12\,p^+ \text{ on } 10\,e^- \;\Rightarrow\; \text{pull per electron}\uparrow \;\Rightarrow\; r_Y < r_X \] Answer: \(\displaystyle r_X > r_Y\); Y has lost its outer shell and the same $\displaystyle 12$ protons hold only $\displaystyle 10$ electrons.

  10. Exercise 36

    Arrange the following elements in increasing order of their atomic radii.
    (a)
    Li, Be, F, N
    (b)
    Cl, At, Br I

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    NCERT’s answer
    (a)
    F < N < Be < Li
    (b)
    Cl < Br < I < At
    Li, Be, N, F lie in Period $\displaystyle 2$: nuclear charge rises left to right while the shell count stays fixed, so pull on the outer electrons grows and radius falls. \[Z \uparrow,\; n \text{ same} \;\Rightarrow\; Z_{\text{eff}} \uparrow \;\Rightarrow\; r \downarrow \] Cl, Br, I, At are Group $\displaystyle 17$ halogens: each step down adds a new shell, so radius grows despite the rising charge. \[n \uparrow \;\Rightarrow\; r \uparrow \] Answer: (a) \(\displaystyle \mathrm{F < N < Be < Li}\) (b) \(\displaystyle \mathrm{Cl < Br < I < At}\)