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NEET Practice Questions (MCQs) with Answers & Solutions

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In which of the folllowing transformations, the bond order has increased and the magnetic behaviour has changed?

$C_{2}^{+} \to C_{2}$

${NO}^{+} \to NO$

$O_{2} \to O_{2}^{+}$

$N_{2} \to N_{2}^{+}$

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Explanation

(a)

(a) $\underset{Paramagnetic}{\underset{BO = 1.5}{C_{2}^{+}}} \overset{}{\to} \underset{Diamagnetic}{\underset{BO = 2}{C_{2}}}$ (b) $\underset{Diamagnetic}{\underset{BO = 3}{{NO}^{+}}} \underset{}{\to} \underset{Paramagnetic}{\underset{BO = 2.5}{NO}}$

(c) $\underset{Paramagnetic}{\underset{BO = 2}{O_{2}}} \overset{}{\to} \underset{Paramagnetic}{\underset{BO = 2.5}{O^{+}}}$ (d) $\underset{Diamagnetic}{\underset{BO = 3}{N_{2}}} \underset{}{\to} \underset{Paramagnetic}{\underset{BO = 2.5}{N_{2}^{+}}}$

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The bond having the highest bond energy is :

C=C

C=S

C=O

P=N

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Explanation

(c) More effective axial and sideways overlapping between atomic orbitals of carbon and those of oxygen atom is higher due to smaller size of oxygen atom. oxygen is more electronegative than oxygen atom, due to these factor CO has highest bond energy .

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H.O.M.O. (Highest Occupied Molecular Orbital) of CO molecular is :

Non-bonding M.O. with slight antibonding character

Non-bonding M.O. with slight bonding character

Pure non-bonding M.O.

None of the above

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Explanation

(a) According to coulsion model, H.O.M.O. (Highest Occupied Molecular Orbital) of CO molecular

is non-bonding M.O. with slight antibonding character.

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Incorrect match is:

Electron geometry Possible molecular shape from respective electron geometery

Tetrahadron - Bent

Trigonal bipyramidal - Traingular planar

Octahedron - Square Pyramidal

Pentagonal bipyramidal - Pentagonal planar

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Explanation

The incorrect match is trigonal bipyramidal electron geometry leading to a triangular planar molecular shape. Trigonal bipyramidal electron geometry should result in a molecular shape that is either trigonal bipyramidal or seesaw-shaped, but not triangular planar.

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In which of the following species central atom is NOT surrounded by exactly 8 valence electrons?

$B {F_{4}}^{-}$

$N C l_{3}$

$P C {l_{4}}^{+}$

$S F_{4}$

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Correctly match is :

$d_{x^{2} - y^{2}}$ atomic orbital $—$ One nodal plane

$P_{y}$ atomic orbital $—$ Two nodal plane

$σ_{p x}$ $— ψ$ (gerade)

$P_{p y}$ $— ψ$ (ungerade)

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Explanation

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Consider the following reaction:

${MX}_{4} + X'_{2} \to {MX}_{4} X'_{2}$

If atomic number of M is 52 and X and X' are halogens and X' is more electronegative than X.

Then choose correct statement regarding given information:

Both X' atoms occupy axial positions which are formed by overlapping of p and d-orbitals only

All $M — X bond lengths are identical in both {MX}_{4} and {MX}_{4} X'_{2} compounds$

Central atom 'M' does not use anyone valence non-axial set of d-orbital in hybridization of final product

Hybridization of central atom 'M' remains same in both reactant and final product.

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Explanation

In MX4X'2, the hybridization of the central atom 'M' involves one set of axial d-orbitals and three sets of p-orbitals. Since no other d-orbitals are used, the correct statement is that the central atom 'M' does not use any non-axial set of d-orbitals in hybridization.

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Which atom can have more than eight valence electrons when it is forming covalent bond?

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Explanation

(d) Due to presence of vacant d-orbital excitation in Cl-atom and it can have more than eight valence electrons whem it is forming hypervalent compound like HCl $O_{4}$

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Which bond is expected to be the least polar?

O-F

P-F

Si-N

B-F

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Explanation

(a) Bond polarity is directly related to difference of electronegativity of bonded atoms .

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Select correct statement(s) :

Acidic strength of $HBr > HCl$ but reverse is true for their reducing property

Basic strength of ${PH}_{3} > {AsH}_{3}$ but reverse is true for their bond angle

Dipole moment of ${CH}_{3} Cl > {CH}_{3} F$ but reverse is true for their $H \hat{C} H$ bond angle

$K_{a_{1}}$ of fumaric acid is higher than maleic acid but reverse is true for their $K_{a_{2}}$

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