Which is the correct sequence in the following properties. For the correct order mark `( T)` and for the incorrect order mark `( F) :` `{:((a)"Acidity order ":SiF_(4) lt SiCl_(4) lt SiBr_(4) lt Sil,,(b)"Melting point":NH_(3) gt SbH_(3) gt AsH_(3) gt PH_(3),,),((c )"Boiling point " :NH_(3) gt SbH_(3) gt AsH_(3) gt PH_(3),,(d)"Dipole moment order ":NH_(3) gt SbH_(3) gt AsH_(3) gt PH_(3),=,5):}`A. `F T F T`B. `T F T F`C. `F F T T `D. `F F T F`

Which is the correct sequence in the following properties. For the cor

1.	Which is the correct sequence in the following properties. For the correct order mark `( T)` and for the incorrect order mark `( F) :` `{:((a)"Acidity order ":SiF_(4) lt SiCl_(4) lt SiBr_(4) lt Sil,,(b)"Melting point":NH_(3) gt SbH_(3) gt AsH_(3) gt PH_(3),,),((c )"Boiling point " :NH_(3) gt SbH_(3) gt AsH_(3) gt PH_(3),,(d)"Dipole moment order ":NH_(3) gt SbH_(3) gt AsH_(3) gt PH_(3),=,5):}`A. `F T F T`B. `T F T F`C. `F F T T `D. `F F T F`
Answer» Correct Answer - A `(a) (F),` As the size of halogen atom increases crowding on Si atom will increases, hence, tendency of attack of Lewis base decreases. `(b)(T), M.P.` of `NH_(3)` is highest due to intermolecular `H-` bonding in it. Next lower `M.P.` will be of `SbH_(3)` follofwed by `AsH_(3)` due to high mol. wt. of `SbH_(3)`. (c) (F) `M.P.` and `B.P.` of increase from `PH_(3)` to `AsH_(3)` due to increase in mol. wt. `NH_(3)` does not follow this trend due to iner molecular H-bonding. (d) (T), Value of bond moment decreases.

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