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NEET NEET Questions — Free MCQ Practice

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Hybridisation of boron in diborane is:

A.

$sp$

B.

${sp}^{2}$

C.

${sp}^{3}$

D.

${sp}^{3} d^{2}$

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Explanation

${sp}^{3}$ -hybridisation, but four bonds are neither linear nor in one plane.

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The blue coloured mineral 'Lapis Lazuli' used as semi precious stone is:

A.

sodium alumino silicate

B.

zinc cobaltate

C.

prussian blue

D.

basic copper carbonate

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Explanation

Lapis Lazuli is a name for sodium alumino silicate.

Lapis lazuli (/ËˆlæpÉªs ËˆlæzjÊŠli, -laÉª/), or lapis for short, is a deep blue metamorphic rock used as a semi-precious stone that has been prized since antiquity for its intense color

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Red liquor is:

A.

${({CH}_{3} COO)}_{3} Al$

B.

$Al {(OH)}_{3}$

C.

${Al}_{2} {({CO}_{3})}_{3}$

D.

${Al}_{2} {({SO}_{4})}_{3}$

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Explanation

${({CH}_{3} COO)}_{3} Al$

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Boron carbide, $B_{4} C$ is widely used for:

A.

making acetylene

B.

making plaster of Paris

C.

as a hardest substance after diamond

D.

making boric acid

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Explanation

$B_{4} C$ is next hardest to diamond.

Diamond	10
Corundum	9
Chrysoberyl	8.5
Topaz	8
Beryl	7.5 to 8
Spinel	7.5 to 8
Euclase	7.5
Zircon	7.5
Cordierite	7 to 7.5
Staurolite	7 to 7.5
Tourmaline	7 to 7.5
Quartz	7
Garnet	6.5 to 7.5
Jadeite	6.5 to 7
Sillimanite

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Which one is explosive?

A.

${PCl}_{5}$

B.

$Pb {({NO}_{3})}_{2}$

C.

${NH}_{4} {NO}_{3} + Al powder$

D.

$C_{6} H_{5} {NO}_{2}$

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Explanation

Ammonium nitrate is not, on its own, an explosive,^[10] but it readily forms explosive mixtures with varying properties when combined with primary explosives such as azides or with fuels such as aluminum powder or fuel oil.

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${BCl}_{3}$ does not exist as dimer but ${BH}_{3}$ exist as dimer $(B_{2} H_{6})$ because:

A.

chlorine is more electronegative than hydrogen

B.

there is $pπ - pπ$ back bonding in ${BCl}_{3}$ but ${BH}_{3}$ does not contain such multiple bonding

C.

large sized chlorine atoms do not fit in between the small boron atoms whereas small sized hydrogen atoms get fitted between boron atoms

D.

none of the above

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Explanation

To form the dimer, the planar molecule has to assume a sort of tetrahedral orientation around the boron atom, as it happens in the case of Al2Cl6, which is the dimer of AlCl3 in the vapour state. But, in all the three boron halides, significant amount of pi-pi interaction occurs between the vacant p-orbital of boron and the filled p-orbitals of the halogens. Such interaction in the three trihalides increases in the order BBr3 < BCl3 < BF3

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Which species does not exist?

A.

${[{BF}_{6}]}^{3 -}$

B.

${[{AlF}_{6}]}^{3 -}$

C.

${[{GaF}_{6}]}^{3 -}$

D.

${[{lnF}_{6}]}^{3 -}$

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Explanation

${[{BF}_{6}]}^{3 -}$ does not exist because boron does not have vacant d-subshells.

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Which of the following imparts green colour to flame?

A.

$B {(OMe)}_{3}$

B.

$Na (OMe)$

C.

$Al {({OBr}_{2})}_{3}$

D.

$Sn {(OH)}_{2}$

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Explanation

Boron compound on heating form $B_{2} O_{3}$ which imparts green flame.

The bead test is a traditional part of qualitative inorganic analysis to test for the presence of certain metals. The oldest one is the borax bead test or blister test. It was introduced by Berzelius in 1812.^[1] Since then other salts were used as fluxing agents, such as sodium carbonate or sodium fluoride. The most important one after borax is microcosmic salt,^[1] which is the basis of the microcosmic salt bead test.^[2]

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Borax bead test is responded by:

A.

divalent metals

B.

heavy metals

C.

light metals

D.

metal which forms coloured metaborates

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Explanation

Metals forming coloured bead can be identified by borax bead test.

The bead test is a traditional part of qualitative inorganic analysis to test for the presence of certain metals. The oldest one is the borax bead test or blister test. It was introduced by Berzelius in 1812.^[1] Since then other salts were used as fluxing agents, such as sodium carbonate or sodium fluoride. The most important one after borax is microcosmic salt,^[1] which is the basis of the microcosmic salt bead test.^[2]

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Reactivity of borazole is greater than that of benzene because:

A.

borazole is non-polar compound

B.

borazole is polar compound

C.

borazole is electron deficient compound

D.

of localized electron in it

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Explanation

Borazole is polar due to $B — N$ bond; benzene is non-polar due to $C — C$ bonds.

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