Which of the following statements is/are correct

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The Cr-O-Cr bond angle in $C r_{2} O_{7}^{2 -}$ is more than Cl-O-Cl bond angle in $C l_{2} O_{7}$ due to less electronegativity of chromium

By heating $K_{2} M n F_{6}$ , fluorine can be prepared

The Cl-O-Cl bond angle in $C l_{2} O_{7}$ is more than that in $C l_{2} O$ due to the larger size of $C l O_{3}$ groups in $C l_{2} O_{7}$ than chlorine atoms in $C l_{2} O$

In aqueous solution hydrazine is weaker base than ammonia

answer is A, B, C.

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Detailed Solution

A) At thought $C l O_{3}$ groups in $C l_{2} O_{7}$ are highly electronegative, their size is apparently responsible for the larger angle in $C l_{2} O_{7} (118.6 º)$ than in $C l_{2} O (110.9 º)$
B) In the $C r_{2} {O_{7}}^{2 -}$ ion highly electropositive Cr atoms allow the central oxygen to attract electrons more strongly than is possible in $C l_{2} O_{7}$ resulting stronger BP-BP repulsions and larger bond angle (126º)
C) In ammonia N is in bond with 3 hydrogens atoms due to which electrons density at N atom is more. In N₂H₄ due to the presence of NH₂ group in the place of one hydrogen atom the electron density at nitrogen in N₂H₄ decreases
D) $M n F_{6}^{2 -}$ act as lews base $M n {F_{6}}^{2 -} + 2 S b F_{5} \to 2 S b F_{6}^{-} + M n F_{4}$ the intermediate can lose F atoms with two of these raicals coupling to form F₂