NEET Practice Questions (MCQs) with Answers & Solutions

Flourine does not form oxyacids because  it is more electronegative than oxygen.

Increase in acid strength indicated decrease in basic strength of the conjugate base. Thus for HOCl<HOCl₂ <HOCl₃ <HOCl₄ , the base strength of anion follows the order $OC{l}^{-}>Cl{O}_2^{-}>Cl{O}_3^{-}>Cl{O}_4^{-}$. The addition of oxygen atom provides a large volume so that charge of the anion is easily accomodated.A large uninegative ion attracts and holds a proton less strongly than a small uninegative ion.

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 

As we move from F^- to I^-, the size of the ion increases and hence electron density decreases from F^- to I^-. As a result,the tendency to attract proton decreases.Thus basic character decreases in the order F^- >Cl^- >Br^- > I^-.

Strongest acid possess smallest pK_a value.

N forms NCl₃, N₂O₅ and Ca₃N₂ . Nitrogen, due to no empty d- orbitals, cannot extend its covalency by more than three and hence do not form NCl₅.Due to its small size and high electronegativity it can accept 3 electrons to form $N^{3-} ion \left(C{a}_3{N}_2\right)$

Because of absence of d-orbitals, oxygen cannot increase its co-ordination number. Oxygen exhibits positive oxidation states of +1 in O₂F₂ and +2 in OF_2.