NEET Practice Questions (MCQs) with Answers & Solutions

No. of hybrid orbitals = Total no. of sigma bonds +lone pair of electrons. The no. of sigma bonds in $N{O}_2^{+},N{O}_3^{-} and N{H}_4^{+}$ are 2,3 and 4 respectively. So the hybridisation is sp, sp² and sp³ respectively.

Due to weak bond H₂Te can furnish hydrogen more readily.The increasing order of acidic nature is

H₂O < H₂S < H₂Se < H₂Te

H₂SO₄ cannot act as reducing agent, because when a substance acts as reducing agent, it is oxidised itself and there is increase in O.N. In H₂SO₄ the O.N. of sulphur is +6,which is maximum.

The acidity decrease in the order SO₃ > P₂O₅ > Al₂O₃ > Na₂O. The Al₂O₃ is atmospheric, Na₂O is basic.

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.

The given element is S.The dioxide of S is SO₂ which can act as reducing agent and oxidising agent.

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.

Because NH₃ reacts with others.

$$\begin{gathered} 2N{H}_3 + H_{2}S{O}_4\to (N{H}_4{)}_{2}S{O}_4 \\ 2N{H}_3+P_2{O}_5 +H_{2}O \to 2N{H}_{4}P{O}_3 \\ CaC{l}_2 +4N{H}_3 \to CaC{l}_{2}4N{H}_3 \end{gathered}$$