Hydrogen MCQ Questions & Answers in Inorganic Chemistry | Chemistry

$$2{F_2} + 2{H_2}O \to 4{H^ + } + 4{F^ - } + {O_2}$$

$${H_2}{O_2}$$  acts as a reducing agent towards powerful oxidising agents.
$${H_2}{O_2} \to 2{H^ + } + {O_2} + 2{e^ - }$$

Elements of group 15 - 17 form electron - rich hydrides. Group 14 elements form electron - precise hydrides.

$$Ca{C_2} + 2{D_2}O \to {C_2}{D_2} + Ca{\left( {OD} \right)_2}$$

$$\eqalign{ & mEq\,\,{\text{of}}\,MnO_4^ - \equiv mEq\,\,{\text{of}}\,{H_2}{O_2} \cr & 10 \times \frac{M}{5} \times 5 \equiv mEq\,\,{\text{of}}\,{{\text{H}}_2}{O_2} \cr & \Rightarrow 10\,mEq \equiv mEq\,\,{\text{of}}\,{H_2}{O_2} \cr & {\text{Weight of}}\,{H_2}{O_2} = 10 \times {10^{ - 3}} \times \frac{{34}}{2} \times 0.17 \cr & = 0.2\,g \cr & {\text{Weight of impure}}\,{H_2}{O_2} \cr & = \frac{{100}}{{85}} \times 0.17 \cr & = 0.2\,g \cr} $$

Heavy water is $${D_2}O$$  hence
number of electrons $$ = 2 + 8 = 10$$
number of protons $$ = 10$$
Atomic mass of $${D_2}O = 4 + 16 = 20$$
hence number of neutron = Atomic mass – number of protons
$$ = 20 - 10 = 10$$

$$\eqalign{ & 1\,N = 5.6\,vol \cr & \Rightarrow {\text{Normalities of two solutions are}}\,1\,N\,{\text{and}}\,2\,N \cr & {\text{Normality of mixture}} \cr & = \frac{{1 \times 1 + 1 \times 2}}{{1 + 1}} = \frac{3}{2} = 1.5\,N \cr & {\text{Volume strength}} \cr & = \frac{3}{2} \times 5.6 \cr & = \frac{{16.8}}{2} \cr & = 8.4\,vol \cr} $$

$$30\,vol$$  of $${H_2}{O_2}$$  means one volume of $${H_2}{O_2}$$  on decom-position will give 30 volume of oxygen.

Polyphosphates like sodium hexametaphosphate (calgon) form soluble complexes with cations like $$C{a^{2 + }}$$ and $$M{g^{2 + }}$$ present in hard water and hence used as water softening agents.
$$N{a_6}{P_6}{O_{18}} \to 2N{a^ + } + N{a_4}{P_6}O_{18}^{2 - }$$
$${M^{2 + }} + N{a_4}{P_6}O_{18}^{2 - } \to $$     $${\left[ {N{a_2}M{P_6}{O_{18}}} \right]^{2 - }} + 2N{a^ + }$$

$$CsH > KH > NaH > LiH$$
The reducing strength of the alkali metal hydrides increases down the group.
$$C{H_4} < N{H_3} < {H_2}O$$
Dipole moment depends upon the difference between the electronegativities of the bonded atoms, and the internuclear separation.
$$P{H_3} < As{H_3} < N{H_3} < Sb{H_3}$$
Boiling point increases with an increase in molecular mass, $$N{H_3}$$  being an exception because of strong hydrogen bonding.
$$Cl < S < N < O < F$$
The strength of hydrogen bonding depends upon the electronegativity and the size. Higher electronegativity and smaller size increases the strength of hydrogen bonding.