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

$$T$$ - is a metal so forms basic hydroxide; high $$pH.$$
$$X$$ - is a non-metal so forms acidic hydroxide; low $$pH.$$

Group 16 elements have valence shell configuration $$n{s^2}n{p^4}.$$  Thus, they are $$p$$ - block elements called chalcogens.

Ina period the value of ionisation potential increases from left to right with breaks where the atoms have some what stable configuration. In this case $$N$$ has half filled stable orbitals. Hence has highest ionisation energy. Thus the correct order is $$B < C < O < N$$
and not as given in option (C)

Electron affinity of $$^9F$$  is less than that of $$^{17}Cl$$

The amount of energy required to remove an electron from the outermost orbit of a gaseous atom is known as ionisation potential. Elements having half-filled or completely filled orbitals are more stable than partially filled orbitals.
In a period from left to right ionisation potential decreases as the atomic number increases. The given elements $$(Be, B, C, N, O)$$    are present in \[\text{II}\] period as
\[\xrightarrow{_{\text{Ionisation}\,\,\text{potential}\,\,\text{increases}}^{Be\,\,\,\,\,\,B\,\,\,\,\,\,\,\,C\,\,\,\,N\,\,\,\,\,O}}\]
But in case of $$Be$$  and $$B, Be$$  has higher ionisation potential due to stable configuration.
$${}_4Be = 1{s^2},2{s^2}$$
$$2{s^2}$$      Stable configuration
$${}_5B = 1{s^2},2{s^2}2{p^1}$$
$$2{p^1}$$      Unstable configuration
In the same way in case of $$N$$  and $$O, N$$  has higher ionisation potential than $$O$$  due to stable configuration
$${}_7N = 1{s^2},2{s^2}2{p^3}$$
$$2{p^3}$$      Stable configuration (half-filled)
$${}_8O = 1{s^2},2{s^2}2{p^4}$$
$$2{p^4}$$      Unstable configuration
So, the correct order of increasing ionisation potential will be
$$B < Be < C < O < N$$

$$\eqalign{ & {\text{Number of moles}} = \frac{1}{{35.5}} \cr & {\text{Given,}}\,1\,eV = 23.06\,kcal\,mo{l^{ - 1}} \cr & 3.7\,eV = 3.7 \times 23.06\,kcal\,mo{l^{ - 1}} \cr & {\text{i}}{\text{.e}}{\text{. }}1{\text{ }}mole{\text{ realease energy}} \cr & = 3.7 \times 23.06\,kcal \cr & \therefore \,\,{\text{Energy released}} \cr & = \frac{1}{{35.5}} \times 3.7 \times 23.06\,kcal \cr & = 2.4\,kcal \cr} $$

Element : $$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^8}4{s^2}\left( {n = 4} \right)$$
Period : As $$n = 4,$$  the element belong to 4^th period
Group : $$ns + \left( {n - 1} \right)d + np = 2 + 8 + 0 = 10$$

In periodic table, the metallic character increases down the group because the ionisation enthalpy decreases down the group and metallic character decreases from left to right because the ionisation enthalpy increases from left to right.

$$CO$$  is a neutral oxide. $$SrO$$  is a basic oxide. $$A{l_2}{O_3}$$  is an amphoteric oxide while $$C{O_2}$$  is an acidic oxide.

$$X$$ has highest $$I{E_1}$$  and $$I{E_2}$$  hence, it is a noble gas. $$Y$$ has low $$I{E_1},$$  but very high $$I{E_2}$$  hence, it is an alkali metal. $$Z$$ has low $$I{E_1}$$  than $$I{E_2}$$  and $$I{E_2}$$  is even lower than $$I{E_2}$$  of alkali metal hence, it is an alkaline earth metal.