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Standard XII

Chemistry

Chemical Properties of Boron Family

Aluminum form...

Question

Aluminum forms $[A I F_{6}]^{3 -}$ but boron does not form $[B F_{6}]^{3 -}$ because:

the atomic size of B is small

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of absence of vacant d-orbital in B atom

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of high I.P of B-atom

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B is non-metal

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Solution

The correct option is B of absence of vacant d-orbital in B atom
Electronic configurations:
$A l^{3 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{0} 3 d^{0}$
$B^{3 +} : 1 s^{2} 2 s^{2} 2 p^{0}$
Therefore, d-orbital is not present in Boron. Hence, $[B F_{6}]^{3 -}$ is diffcult to form.

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Similar questions

Q. Assertion :Aluminium forms

[A l F_{6}]^{3 -}

but B does not form

[B F_{6}]^{3 -}

Reason:

B F_{3}

on hydrolysis gives

H B F_{4}

Q. Assertion :

A l

forms

{[A l F_{6}]}^{3 -}

but

B

does not form

{[{B F}_{6}]}^{3 -}

Reason:

B

does not react with fluorine.

Read the above assertion and reason and choose the correct option regarding it.

Q. Explain the following:
(i) Gallium has higher ionization enthalpy than aluminum.
(ii) Boron does not exist as

B^{3 +}

ion.
(iii) Aluminum forms

[A l F_{6}]^{3 -}

ion but boron does not form

[B F_{6}]^{3 -}

ion.
(iv)

P b X_{2}

is more stable than

P b X_{4}

(v)

P b^{4 +}

acts as an oxidizing agent, but

S n^{2 +}

acts as a reducing agent.
(vi) Electron gain enthalpy of chlorine is more negative as compared to fluorine.
(vii)

T I (N O_{3})_{3}

acts as an oxidizing agent.
(viii) Carbon shows catenation property but lead does not.
(ix)

B F_{3}

does not hydrolyze.
(x) Why does the element silicon, not form a graphite-like structure whereas carbon does?

Q. Why does boron not form

B^{3 +}

ion?

Q. Boron does not from

B^{3 +}

ions whereas

A l

forms

A l^{3 +} i o n s

. This because :

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Aluminum forms [AIF6]3− but boron does not form [BF6]3− because:

The correct option is B of absence of vacant d-orbital in B atomElectronic configurations:Al3+:1s22s22p63s23p03d0B3+:1s22s22p0Therefore, d-orbital is not present in Boron. Hence, [BF6]3− is diffcult to form.

Aluminum forms $[A I F_{6}]^{3 -}$ but boron does not form $[B F_{6}]^{3 -}$ because:

The correct option is B of absence of vacant d-orbital in B atom
Electronic configurations:
$A l^{3 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{0} 3 d^{0}$
$B^{3 +} : 1 s^{2} 2 s^{2} 2 p^{0}$
Therefore, d-orbital is not present in Boron. Hence, $[B F_{6}]^{3 -}$ is diffcult to form.