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Standard XII

Chemistry

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Match the ele...

Question

Match the elements with the highest exhibited property:

Column I Column II

P Cs 1 atomic size

Q F 2 Electron affinity

R Cl 3 Electronegativity

S Ar 4 Metallic character

P - 4, Q - 3, R - 2, S - 1

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P - 4, Q - 3, R - 1, S - 2

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P - 1, Q - 2, R - 3, S - 4

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P - 3, Q - 2, R - 4, S - 1

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Solution

The correct option is A $P - 4, Q - 3, R - 2, S - 1$
$C s$ has the highest metallic character.
Fluorine ( $F$ ) is the most electronegative element
Chlorine ( $C l$ ) has the highest electron affinity
$A r$ has van der Waals radius and hence it has greatest atomic radius.

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Q. Match the elements with the highest exhibited property:

	Column I		Column II
P	Cs	1	atomic size
Q	F	2	Electron affinity
R	Cl	3	Electronegativity
S	Ar	4	Metallic character

Q. Match the increasing orders given in column I with the property(ies) given in column II.

Column I	Column II
(A) $N a^{+} < F^{-} < O^{2 -} < N^{3 -}$	(p) electronegativity
(B) $L i^{+}$ < $N a^{+}$ < $K^{+}$ < $R b^{+}$ < $C s^{+}$	(q) Ionisation energy
(C) $O < S < F < C l$	(r) Size
(D) $C l^{-} < K^{+} < C a^{2 +} < S c^{3 +}$	(s) Electron affinity

Q. Explain:
(i) Larger the atomic size, more metallic is the element.
(ii) Halogens have high electron affinity.
(iii) Size of atom changes when it loses or gain electron.
(iv) K is more reactive than Li.
(v) Electronegativity of Cl is higher than S.
(vi) Group 17 elements are non-metals, while group 1 elements are metals.

Arrange the following as per instructions given in the brackets.

(a) Mg. Cl, Na, S, Si (decreasing order of atomic size)

(b) Cs, Na, Li, K, Rb (increasing metallic character)

(d) Cl, F, Br, I (increasing electron affinity)

(e) Cs, Na, Li, K, Rb (decreasing electronegativity)

Column-I: Property; Column-II: Variation of property; Column-III: Magnitude
$\begin{matrix} Column-I & Column-II & Column-III (I) & Electron a affinity (E A_{1}) & (i) & Decreases along the period & (P) & Highest in halogen in their respective periods (I I) & Ionization energy (I E_{1}) & (i i) & Directly proportional to Z_{e f f} & (Q) & Highest in noble gases in their respective periods (I I I) & Electronegativity & (i i i) & Decreases down the group & (R) & Highest in alkali metals in their respective periods (I V) & Electropositive character & (i v) & Inversly proportional to size & (S) & Moderate in noble gases in their respective periods \end{matrix}$
The only correct combination for the energy required to knock out most loosely bounded electron from isolated gaseous atom is:

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Standard XII Chemistry

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Match the elements with the highest exhibited property: Column I Column II P Cs 1 atomic size Q F 2 Electron affinity R Cl 3 Electronegativity S Ar 4 Metallic character

The correct option is A P−4,Q−3,R−2,S−1Cs has the highest metallic character. Fluorine (F) is the most electronegative element Chlorine (Cl) has the highest electron affinity Ar has van der Waals radius and hence it has greatest atomic radius.

Match the elements with the highest exhibited property:

Column I Column II

P Cs 1 atomic size

Q F 2 Electron affinity

R Cl 3 Electronegativity

S Ar 4 Metallic character

The correct option is A $P - 4, Q - 3, R - 2, S - 1$
$C s$ has the highest metallic character.
Fluorine ( $F$ ) is the most electronegative element
Chlorine ( $C l$ ) has the highest electron affinity
$A r$ has van der Waals radius and hence it has greatest atomic radius.