Which is the correct sequence in the following properties? For the correct order mark(T) and for the incorrect order mark (F):
(a) Lewis acidity order : $S i F_{4} < S i C l_{4} < S i B r_{4} < S i I_{4}$
(b) Melting point: $N H_{3} > S b H_{3} > A s H_{3} > P H_{3}$
(c) Boiling point : $N H_{3} > S b H_{3} > A s H_{3} > P H_{3}$
(d) Dipole moment order : $N H_{3} > S b H_{3} > A s H_{3} > P H_{3}$

FTFT

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TFTF

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FFTT

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FFTF

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Solution

The correct option is A FTFT
(1) Melting points of members of group 15 follows the order $P H_{3} < A s H_{3} < S b H_{3} < N H_{3}$
because of intermolecular H-bonding, $N H_{3}$ has the highest melting point. As we move down the group from $P H_{3} t o S b H_{3}$
because of intermolecular H bonding, Nitrogen hydroxide has the highest melting point will increase due to increase in van der walls force of attraction
(2) Boiling point the member will increase regularly as use more down the group from $P H_{3}$ to $B i H_{3}$ will have the highest boiling point than $P H_{3}$ , because of intermolecular hydrogen bonding but for the rest of the members vanderwalls force of attraction increase as molecular mass increase
as we move down the group 15, the electronegativity of the atoms decreases because of increases in their size the dipole moment is the product of change and distance between them. how the size of the atom also increases down the group also considering the fact that nitrogen is the most electronegative elements among them.

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Similar questions

1. S b H_{3} > N H_{3} > A s H_{3} > P H_{3}

2. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

3. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

4. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

1. S b H_{3} > N H_{3} > A s H_{3} > P H_{3}

2. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

3. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

4. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

N H_{3}, P H_{3}, A s H_{3}, S b H_{3}

N H_{3}, P H_{3}, A s H_{3}, S b H_{3}

N H_{3}, P H_{3}, A s H_{3}

S b H_{3}

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