Chemistry MCQs for NEET — Practice Questions with Answers

The acidic strength of oxides decreases on moving down the group.Moreover, pentoxides are more acidic than trioxides. As the atomic no. of an element in 15 group increases, the acidic character decreases.

So,  P₄O₁₀ > As₄O₆ and P₄O₁₀ > As₄O₁₀ . Due to increase in O.N. of an element acidic nature increases.Thus decreasing order of acidic strength is P₄O₁₀ > P₄O₆  > As₄O₁₀ > As₄O₆ 

The stability decrease in the order $I_3^{-}$ > $B{r}_3^{-}$ > $C{l}_3^{-}$ > $F_3^{-}$ 

In fact no $F_3^{-}$ compounds are known,because flourine has no available d-orbitals and therefore can not expand its octet.

Multiple bonds formation tendency with carbon and nitrogen decreases from sulphur to tellurium.

CS₂ (S=C=S) is moderately stable,

CSe₂ (Se=C=Se) decomposes readily whereas,

CTe₂ (Te=C=Te) does not exist

HClO is the weakest acid,when compared to the order acids listed.pK_a =8 for HClO and for HClO₄ ,pK_a =1.

The oxidation state in $N_{2}O <NO<N_2{O}_3<N_2{O}_4<N_2{O}_5$ dereases in the order +1<+2<+3<+4<+5.

In fact, N₂O and NO are neutral oxides.

N is more electroneative than I.The difference of electronegativity is largest in the case of atoms N and I. The dipole moment of the bonded atoms adds to that of the lone pair and thus polarity increases.

The ionic charcter is expected to increase as the electronegativity of the halogen increases. Thus due to the highest electronegativity difference in H and F, the H-F bond should be most ionic.

% of ionic character=16×ΔEN+3.5×(ΔEN)²

H-I bond being the weakest gives the hydrogen readily. HI is also the strongest acid because of its least bond dissociation energy.The acidic  character thus follows the order HI>HBr>HCl>HF.

The stability of the highest oxidation states decreases down the group and Bi does not form a pentoxide.Moreover higher oxidation states are more acidic than lower oxidation states.

Flourine reacts with dilute NaOH to form fluoride and oxygen fluoride

$$\begin{gathered} 2F_2 +2NaOH \to 2NaF +O{F}_2+H_{2}O \\ with hot and conc. NaOH, O_2 is liberated \\ 2F_2 +4NaOH\to 4NaF +O_2 +2H_{2}O \end{gathered}$$

tendency to get oxidise oxyacids of chlorine follows the order :

HClO >HClO₂ > HClO₃ > HClO₄