Classification of Elements and Periodicity in Properties MCQ Questions & Answers in Inorganic Chemistry | Chemistry

Key Idea Across a period, increasing nuclear charge outweighs the shielding, hence the outermost electrons are held more and more tightly and ionisation energy. increases across a period while as we move down a group increase in shielding outweighs the increasing nuclear charge and the removal of the outermost electron required less energy down a group.

Electronic configuration	Group
$$\left[ {Ne} \right]3{s^2}3{p^3}$$	$${\text{V}}$$
$$\left[ {Ne} \right]3{s^2}3{p^2}$$	$${\text{IV}}$$
$$\left[ {Ar} \right]3{d^{10}},4{s^2}4{p^3}$$	$${\text{V}}$$
$$\left[ {Ne} \right]3{s^2}3{p^1}$$	$${\text{III}}$$

Since, ionisation energy increases in a period and decreases in a group, $$\left[ {Ne} \right]3{s^2}3{p^3}$$   configuration has the highest ionisation energy among the given elements.

Element $$X$$ has atomic number 19. Its valency will be one. Hence, the formula of its oxide will be $${X_2}O.$$
$$Z = 19;1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}$$

The amount of energy required to remove an electron from unipositive ion is referred as second ionisation potential.
In $$Ti, V, Cr$$   and $$Mn,$$  generally second ionisation energy increases with increase in atomic number but second ionisation potential of $$Cr$$  is greater than that of $$Mn$$  due to the presence of exactly half-filled $$d$$-subshell in $$Cr.$$
Thus, the order of second ionisation enthalpy is $$Cr > Mn > V > Ti$$

The effective nuclear charge decreases when an anion is formed due to increase in number of electrons.

TIPS/Formulae :
(i) Hydrogen bonding increases the boiling point.
(ii) Hydrogen bonds are formed in compounds having $$F$$ or $$O$$ or $$N$$ with hydrogen
$$S,\,\,Se,\,\,Te$$   cannot undergo hydrogen bond formation because of their larger size and lower electronegativity values.

(A) All of these are isoelectronic ions with decreasing nuclear charge and increasing ionic size.
$$A{l^{3 + }}\left( {Z = 13} \right) < M{g^{2 + }}\left( {Z = 12} \right)$$       $$ < N{a^ + }\left( {Z = 11} \right) < {F^ - }\left( {Z = 9} \right)$$
(B) $$I{E_1}$$  of these elements are in the order : $$B < C < O < N.$$     $$I{E_1}$$  of $$N\left( {1{s^2}2{s^2}2{p^3}} \right)$$   is more than $$O\left( {1{s^2}2{s^2}2{p^4}} \right)$$   due to extra stable half-filled $$2p$$ -subshell in the valence shell. Thus, the given order is not correct.
(C) Electron gain enthalpies of these elements are $$I\left( { - 295\,kJ\,mo{l^{ - 1}}} \right) < $$     $$Br\left( { - 325\,kJ\,mo{l^{ - 1}}} \right) < $$     $$F\left( { - 328\,kJ\,mo{l^{ - 1}}} \right) < $$     $$Cl\left( { - 349\,kJ\,mo{l^{ - 1}}} \right)$$
(D) Metallic radius increases down the group in I^st group.

$$Li, Be, B$$   and $$C$$  are present in II^nd period. In a period from left to right ionisation potential increases.
\[\xrightarrow[Li\,\,\,Be\,\,\,\,B\,\,\,\,C]{\text{Ionisation}\,\,\text{potention}\,\,\text{increase}}\]
* But in case of $$Be$$  and $$B, Be$$  has higher ionisation potential than $$B$$  due to stable configuration of $$Be.$$
$${}_4Be = 1{s^2},2{s^2}$$
 Stable configuration ( due to fully- filled orbital )
$${}_5B = 1{s^2},2{s^2}2{p^1}$$
Unstable configuration
So, the correct order of ionisation potential of given elements is
$$Li < B < Be < C$$

Acidic character of oxide Non-metallic nature of element.
Non-metallic character increases along the period. Hence order of acidic character is
$$C{l_2}{O_7} > S{O_2} > {P_4}{O_{10}}.$$

Electronegativity of an atom $${\text{ = }}\frac{{IE + EA}}{{2 \times 62.5}}$$
where $$IE$$  and $$EA$$  are in $$k\,cal\,mo{l^{ - 1}}.$$

$$N{a_2}O\,\,\left( {{\text{basic}}} \right),S{O_2}\,{\text{and}}\,{B_2}{O_3}\,\left( {{\text{acidic}}} \right)\,{\text{and}}\,ZnO\,\,{\text{is}}\,{\text{amphoteric}}{\text{.}}$$