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NCERT Exemplar · Class 10 Science Periodic Classification of Elements

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Long Answer Questions 42–52 (part 6 of 6)

  1. Exercise 42

    An element is placed in 2nd Group and 3rd Period of the Periodic Table, burns in presence of oxygen to form a basic oxide.
    (a)
    Identify the element
    (b)
    Write the electronic configuration
    (c)
    Write the balanced equation when it burns in the presence of air
    (d)
    Write a balanced equation when this oxide is dissolved in water
    (e)
    Draw the electron dot structure for the formation of this oxide

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    NCERT’s answer
    (a)
    Magnesium (Mg)
    (b)
    K, L, M
    $\displaystyle 2$, $\displaystyle 8$, $\displaystyle 2$
    (c)
    \(\displaystyle 2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{MgO}(\mathrm{s})\)
    (d)
    \(\displaystyle \mathrm{MgO}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Mg}(\mathrm{OH})_2(\mathrm{aq})\)
    (e)
    NCERT_Solution_Class10_Science_Exemplar_Ch5_Q42_ncert
    (a)
    Group $\displaystyle 2$, Period $\displaystyle 3$ \(\displaystyle \Rightarrow\) Magnesium (Mg).
    (b)
    Electronic configuration: \(\displaystyle 2, 8, 2\).
    (c)
    \[2\mathrm{Mg} + \mathrm{O_2} \xrightarrow{\Delta} 2\mathrm{MgO} \]
    (d)
    \[\mathrm{MgO}(s) + \mathrm{H_2O}(l) \rightarrow \mathrm{Mg(OH)_2}(aq) \]
    (e)
    Mg transfers its $\displaystyle 2$ valence electrons to O, giving \(\displaystyle \mathrm{Mg^{2+}}\) and \(\displaystyle \mathrm{O^{2-}}\), each with a stable octet.
    Answer: Mg; \(\displaystyle 2,8,2\); \(\displaystyle 2\mathrm{Mg}+\mathrm{O_2}\rightarrow 2\mathrm{MgO}\); \(\displaystyle \mathrm{MgO}+\mathrm{H_2O}\rightarrow \mathrm{Mg(OH)_2}\); ionic \(\displaystyle \mathrm{Mg^{2+}O^{2-}}\).

  2. Exercise 43

    An element X (atomic number 17\displaystyle 17) reacts with an element Y (atomic number 20\displaystyle 20) to form a divalent halide.
    (a)
    Where in the periodic table are elements X and Y placed?
    (b)
    Classify X and Y as metal (s), non-metal (s) or metalloid (s)
    (c)
    What will be the nature of oxide of element Y? Identify the nature of bonding in the compound formed
    (d)
    Draw the electron dot structure of the divalent halide

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    NCERT’s answer
    (a)
    X belongs to Group $\displaystyle 17$ and \(\displaystyle 3^{\text {rd }}\) period
    Y belongs to Group $\displaystyle 2$ and \(\displaystyle 4^{\text {th }}\) period
    (b)
    X — Non-metal and Y — Metal
    (c)
    Basic oxide; Ionic bonding
    (d)
    NCERT_Solution_Class10_Science_Exemplar_Ch5_Q43_ncert
    (a)
    X (\(\displaystyle Z=17\), \(\displaystyle \mathrm{Cl}\)): Group $\displaystyle 17$, Period 3. Y (\(\displaystyle Z=20\), \(\displaystyle \mathrm{Ca}\)): Group $\displaystyle 2$, Period 4.
    (b)
    X = non-metal; Y = metal.
    (c)
    Oxide of Y is \(\displaystyle \mathrm{CaO}\), a basic oxide. Bonding in the halide \(\displaystyle \mathrm{CaCl_2}\) is ionic (electrovalent) — Ca loses $\displaystyle 2$ electrons, one to each Cl.
    (d)
    Answer: X = Cl (Group $\displaystyle 17$, Period $\displaystyle 3$, non-metal); Y = Ca (Group $\displaystyle 2$, Period $\displaystyle 4$, metal); \(\displaystyle \mathrm{CaO}\) is basic; \(\displaystyle \mathrm{CaCl_2}\) bonding is ionic.

  3. Exercise 44

    Atomic number of a few elements are given below
    10\displaystyle 10, 20\displaystyle 20, 7\displaystyle 7, 14\displaystyle 14
    (a)
    Identify the elements
    (b)
    Identify the Group number of these elements in the Periodic Table
    (c)
    Identify the Periods of these elements in the Periodic Table
    (d)
    What would be the electronic configuration for each of these elements?
    (e)
    Determine the valency of these elements

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    (a)
    Elements— Neon (Ne), Calcium (Ca), Nitrogen (N), Silicon (Si)
    (b)
    Group— $\displaystyle 18$, $\displaystyle 2$, $\displaystyle 15$, $\displaystyle 14$
    (c)
    Period— $\displaystyle 2$, $\displaystyle 4$ , $\displaystyle 2$, $\displaystyle 3$
    (d)
    Electron configuration— $\displaystyle (2, 8)$; ($\displaystyle 2$, $\displaystyle 8$, $\displaystyle 8$, $\displaystyle 2$); $\displaystyle (2, 5)$; ($\displaystyle 2$, $\displaystyle 8$, $\displaystyle 4$)
    (e)
    Valency— $\displaystyle 0$, $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$
    (a)
    \(\displaystyle Z=10\): \(\displaystyle \mathrm{Ne}\); \(\displaystyle Z=20\): \(\displaystyle \mathrm{Ca}\); \(\displaystyle Z=7\): \(\displaystyle \mathrm{N}\); \(\displaystyle Z=14\): \(\displaystyle \mathrm{Si}\).
    (b)
    Groups: Ne = $\displaystyle 18$, Ca = $\displaystyle 2$, N = $\displaystyle 15$, Si = 14.
    (c)
    Periods: Ne = $\displaystyle 2$, Ca = $\displaystyle 4$, N = $\displaystyle 2$, Si = 3.
    (d)
    Configurations: Ne \(\displaystyle 2,8\); Ca \(\displaystyle 2,8,8,2\); N \(\displaystyle 2,5\); Si \(\displaystyle 2,8,4\).
    (e)
    Valency: Ne = \(\displaystyle 0\) (complete octet, inert); Ca = \(\displaystyle 2\); N = \(\displaystyle 3\); Si = \(\displaystyle 4\).
    Answer: Ne (Group $\displaystyle 18$, Period $\displaystyle 2$, valency $\displaystyle 0$); Ca (Group $\displaystyle 2$, Period $\displaystyle 4$, valency $\displaystyle 2$); N (Group $\displaystyle 15$, Period $\displaystyle 2$, valency $\displaystyle 3$); Si (Group $\displaystyle 14$, Period $\displaystyle 3$, valency $\displaystyle 4$).

  4. Exercise 45

    Complete the following cross word puzzle (Figure 5.1\displaystyle 5.1)
    Across:
    (1)
    An element with atomic number 12.
    (3)
    Metal used in making cans and member of Group 14.
    (4)
    A lustrous non-metal which has 7\displaystyle 7 electrons in its outermost shell.
    Down:
    (2)
    Highly reactive and soft metal which imparts yellow colour when subjected to flame and is kept in kerosene.
    (5)
    The first element of second Period
    (6)
    An element which is used in making fluorescent bulbs and is second member of Group 18\displaystyle 18 in the Modern Periodic Table
    (7)
    A radioactive element which is the last member of halogen family.
    (8)
    Metal which is an important constituent of steel and forms rust when exposed to moist air.
    (9)
    The first metalloid in Modern Periodic Table whose fibres are used in making bullet-proof vests
    NCERT_Question_Class10_Science_Exemplar_Ch5_Q45

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    NCERT_Solution_Class10_Science_Exemplar_Ch5_Q45_ncert
    Across — ($\displaystyle 1$) \(\displaystyle \mathrm{Mg}\) Magnesium; ($\displaystyle 3$) \(\displaystyle \mathrm{Sn}\) Tin; ($\displaystyle 4$) \(\displaystyle \mathrm{I}\) Iodine. Down — ($\displaystyle 2$) \(\displaystyle \mathrm{Na}\) Sodium; ($\displaystyle 5$) \(\displaystyle \mathrm{Li}\) Lithium; ($\displaystyle 6$) \(\displaystyle \mathrm{Ne}\) Neon; ($\displaystyle 7$) \(\displaystyle \mathrm{At}\) Astatine; ($\displaystyle 8$) \(\displaystyle \mathrm{Fe}\) Iron; ($\displaystyle 9$) \(\displaystyle \mathrm{B}\) Boron.Answer: $\displaystyle 1$ Mg, $\displaystyle 2$ Na, $\displaystyle 3$ Sn, $\displaystyle 4$ I, $\displaystyle 5$ Li, $\displaystyle 6$ Ne, $\displaystyle 7$ At, $\displaystyle 8$ Fe, $\displaystyle 9$ B.

  5. Exercise 46

    (a)
    In this ladder (Figure 5.2\displaystyle 5.2) symbols of elements are jumbled up. Rearrange these symbols of elements in the increasing order of their atomic number in the Periodic Table.
    (b)
    Arrange them in the order of their group also.
    NCERT_Question_Class10_Science_Exemplar_Ch5_Q46

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    (a)
    H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca
    (b)
    Group $\displaystyle 1$ — H, Li, Na, K
    Group $\displaystyle 2$ — Be, Mg, Ca
    Group $\displaystyle 13$ — B, Al
    Group $\displaystyle 14$ — C, Si
    Group $\displaystyle 15$ — N, P
    Group $\displaystyle 16$ — O, S
    Group $\displaystyle 17$ — F, Cl
    Group $\displaystyle 18$ — He, Ne, Ar
    (a)
    By increasing atomic number ($\displaystyle 1$–$\displaystyle 20$):
    \[\mathrm{H,\ He,\ Li,\ Be,\ B,\ C,\ N,\ O,\ F,\ Ne,\ Na,\ Mg,\ Al,\ Si,\ P,\ S,\ Cl,\ Ar,\ K,\ Ca} \]
    (b)
    By group: $\displaystyle 1$ — H, Li, Na, K; $\displaystyle 2$ — Be, Mg, Ca; $\displaystyle 13$ — B, Al; $\displaystyle 14$ — C, Si; $\displaystyle 15$ — N, P; $\displaystyle 16$ — O, S; $\displaystyle 17$ — F, Cl; $\displaystyle 18$ — He, Ne, Ar.
    Answer: (a) \(\displaystyle \mathrm{H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca}\) (\(\displaystyle Z=1\) to \(\displaystyle 20\)); (b) Group $\displaystyle 1$: H, Li, Na, K; $\displaystyle 2$: Be, Mg, Ca; $\displaystyle 13$: B, Al; $\displaystyle 14$: C, Si; $\displaystyle 15$: N, P; $\displaystyle 16$: O, S; $\displaystyle 17$: F, Cl; $\displaystyle 18$: He, Ne, Ar.

  6. Exercise 47

    Mendeléev predicted the existence of certain elements not known at that time and named two of them as Eka-silicon and Eka-aluminium.
    (a)
    Name the elements which have taken the place of these elements
    (b)
    Mention the group and the period of these elements in the Modern Periodic Table.
    (c)
    Classify these elements as metals, non-metals or metalloids
    (d)
    How many valence electrons are present in each one of them?

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    NCERT’s answer
    (a)
    Germanium (Ge) and Gallium (Ga)
    (b)
    Group $\displaystyle 14$; Period $\displaystyle 4$ and Group $\displaystyle 13$; Period $\displaystyle 4$
    (c)
    Ge — Metalloid; Ga — Metal
    (d)
    Ga — $\displaystyle 3$ Ge —$\displaystyle 4$
    (a)
    Eka-silicon \(\displaystyle \Rightarrow\) Germanium (\(\displaystyle \mathrm{Ge}\)); Eka-aluminium \(\displaystyle \Rightarrow\) Gallium (\(\displaystyle \mathrm{Ga}\)).
    (b)
    Ge: Group $\displaystyle 14$, Period 4. Ga: Group $\displaystyle 13$, Period 4.
    (c)
    Ge is a metalloid; Ga is a metal.
    (d)
    Valence electrons: Ge = \(\displaystyle 4\); Ga = \(\displaystyle 3\).
    Answer: Eka-silicon = Ge (Group $\displaystyle 14$, Period $\displaystyle 4$, metalloid, $\displaystyle 4$ valence e\(\displaystyle ^-\)); Eka-aluminium = Ga (Group $\displaystyle 13$, Period $\displaystyle 4$, metal, $\displaystyle 3$ valence e\(\displaystyle ^-\)).

  7. Exercise 48

    (a)
    Electropositive nature of the element(s) increases down the group and decreases across the period
    (b)
    Electronegativity of the element decreases down the group and increases across the period
    (c)
    Atomic size increases down the group and decreases across a period (left to right)
    (d)
    Metallic character increases down the group and decreases across a period.
    On the basis of the above trends of the Periodic Table, answer the following about the elements with atomic numbers 3\displaystyle 3 to 9.
    (a)
    Name the most electropositive element among them
    (b)
    Name the most electronegative element
    (c)
    Name the element with smallest atomic size
    (d)
    Name the element which is a metalloid
    (e)
    Name the element which shows maximum valency.

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    NCERT’s answer
    (a)
    Lithium
    (b)
    Fluorine
    (c)
    Fluorine
    (d)
    Boron
    (e)
    Carbon
    Elements $\displaystyle 3$–$\displaystyle 9$ form Period $\displaystyle 2$: Li, Be, B, C, N, O, F.
    (a)
    Electropositive character falls across a period, so it peaks at the left end: Lithium.
    (b)
    Electronegativity rises across a period, so it peaks at the right end: Fluorine.
    (c)
    Atomic size falls across a period, so it is smallest at the right end: Fluorine.
    (d)
    The element with mixed metal/non-metal character in this row is the metalloid Boron.
    (e)
    Bonding capacity is greatest at the row's middle ($\displaystyle 4$ electrons available for bonds): Carbon.

  8. Exercise 49

    An element X which is a yellow solid at room temperature shows catenation and allotropy. X forms two oxides which are also formed during the thermal decomposition of ferrous sulphate crystals and are the major air pollutants.
    (a)
    Identify the element X
    (b)
    Write the electronic configuration of X
    (c)
    Write the balanced chemical equation for the thermal decomposition of ferrous sulphate crystals?
    (d)
    What would be the nature (acidic/ basic) of oxides formed?
    (e)
    Locate the position of the element in the Modern Periodic Table.

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    NCERT’s answer
    (a)
    Element X is sulphur (atomic no. $\displaystyle 16$)
    (b)
    K, L, M
    $\displaystyle 2$, $\displaystyle 8$, $\displaystyle 6$
    (c)
    \(\displaystyle 2 \mathrm{FeSO}_4(\mathrm{~s}) \xrightarrow{\text { Heat }} \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+\mathrm{SO}_2(\mathrm{~g})+\mathrm{SO}_3(\mathrm{~g})\)
    (d)
    Acidic
    (e)
    3rd period, group $\displaystyle 16$
    (a)
    X = Sulphur (S).
    (b)
    Atomic number $\displaystyle 16$ → configuration \(\displaystyle 2, 8, 6\).
    (c)
    \[2\,\mathrm{FeSO_4} \xrightarrow{\Delta} \mathrm{Fe_2O_3} + \mathrm{SO_2} + \mathrm{SO_3} \]
    (d)
    SO₂ and SO₃ are non-metal oxides, hence acidic.
    (e)
    $\displaystyle 6$ valence electrons → Group $\displaystyle 16$; $\displaystyle 3$ shells → Period $\displaystyle 3$.

  9. Exercise 50

    An element X of group 15\displaystyle 15 exists as diatomic molecule and combines with hydrogen at 773\displaystyle 773 K in presence of the catalyst to form a compound, ammonia which has a characteristic pungent smell.
    (a)
    Identify the element X. How many valence electrons does it have?
    (b)
    Draw the electron dot structure of the diatomic molecule of X. What type of bond is formed in it?
    (c)
    Draw the electron dot structure for ammonia and what type of bond is formed in it?

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    NCERT’s answer
    (a)
    Nitrogen (atomic no. $\displaystyle 7$)
    $\displaystyle 2,5$; it has $\displaystyle 5$ valence electrons
    (b)
    triple covalent bonds
    (c)
    $\displaystyle 3$ single covalent bonds
    NCERT_Solution_Class10_Science_Exemplar_Ch5_Q50_ncert
    NCERT_Solution_Class10_Science_Exemplar_Ch5_Q50_ncert_2
    (a)
    X = Nitrogen (N), $\displaystyle 5$ valence electrons.
    (b)
    Each N atom needs $\displaystyle 3$ more electrons for an octet, so the two atoms share three electron pairs — a triple covalent bond.
    (c)
    In ammonia, N shares one electron pair with each H atom (three single covalent bonds) and keeps one lone pair.

  10. Exercise 51

    Which group of elements could be placed in Mendeléev's Periodic Table without disturbing the original order? Give reason.

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    NCERT’s answer
    Noble gases According to Mendeleev's classification, the properties of elements are the periodic function of their atomic masses and there is a periodic recurrence of elements with similar physical and chemical properties. Noble gas being inert, could be placed in a separate group without disturbing the original order.
    Noble gases (He, Ne, Ar, Kr, Xe, Rn). They were unknown when Mendeléev built his table, and their extremely low reactivity (valency zero) sets them apart from every listed group. A new column could be inserted for them at the end of each period, so none of the $\displaystyle 63$ already-placed elements had to move.

  11. Exercise 52

    Give an account of the process adopted by Mendeléev for the classification of elements. How did he arrive at "Periodic Law"?

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    NCERT’s answer
    (Hint— $\displaystyle 63$ elements were known.)
    Compounds of these elements with oxygen and hydrogen were studied (formation of oxides and hydrides)
    Elements with similar properties were arranged in a group
    Mendeléev observed that elements were automatically arranged in the order of increasing atomic masses.
    Mendeléev arranged the $\displaystyle 63$ then-known elements in increasing order of atomic mass. To judge similarity he compared the formulae and properties of each element's oxides and hydrides, since oxygen and hydrogen combine with most elements. Elements with similar properties went into the same vertical group; successive elements of rising mass formed a period. Where no known element fitted a slot, he left a gap and predicted its properties from its neighbours (eka-aluminium and eka-silicon, later Gallium and Germanium). Similar properties recurred at regular intervals on this arrangement, so he stated the Periodic Law.Answer: The properties of elements are a periodic function of their atomic masses.