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Test: Hybridisation - NEET MCQ


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22 Questions MCQ Test Chemistry Class 11 - Test: Hybridisation

Test: Hybridisation for NEET 2024 is part of Chemistry Class 11 preparation. The Test: Hybridisation questions and answers have been prepared according to the NEET exam syllabus.The Test: Hybridisation MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Hybridisation below.
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Test: Hybridisation - Question 1

Direction (Q. Nos. 1-11) This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q.  Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is 

Detailed Solution for Test: Hybridisation - Question 1

Test: Hybridisation - Question 2

Which of the following statement is incorrect regarding NO2 molecule?

Detailed Solution for Test: Hybridisation - Question 2

NO2 molecule has unpaired electrons which are responsible for its brown colour and paramagnetic behavior.

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Test: Hybridisation - Question 3

Molecule MX3 (atomic number M < 21) has zero dipole moment, the sigma bonding orbitals used by M are

Detailed Solution for Test: Hybridisation - Question 3

Molecule has zero dipole moment, thus it does not have lone pair. Hence, non-polar.

Thus, (M — X) bonds indicate sp2-hybridisation.

Test: Hybridisation - Question 4

bond in between (C2— C3) vinyl acetylene is formed by ...... overlapping

Detailed Solution for Test: Hybridisation - Question 4

Vinyl acetylene is

bond are at equal position, numbering of C-chains based on IUPAC nomenclature is done from side.

Test: Hybridisation - Question 5

Hydrogen bonds are formed in many compounds e.g., H2O, HF, NH3 . The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of above compounds is :

Detailed Solution for Test: Hybridisation - Question 5

H2​O>HF>NH3​

Strength of hydrogen bonding depends on the size and electronegativity of the atom. 

Smaller the size of the atom, greater is the electronegativity and hence stronger is the H−bonding. Thus, the order of strength of H-bonding is H...F>H...O>H...N. 

But each HF molecule is linked only to two other HF molecules while each H2O molecule is linked to four other H2​O molecules through H−bonding.

Hence, the decreasing order of boiling points is H2​O>HF>NH3​.

Test: Hybridisation - Question 6

Considering the state of hybridisation of C-atoms,select the molecule among the following which is linear

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Test: Hybridisation - Question 7

In which one of the following species the central atom has the type of hybridisation which si not the same as that present in the other three? 

Detailed Solution for Test: Hybridisation - Question 7

Molecules having the same number of hybrid orbitals, have same hybridisation and number of hybrid oebitals.

where,

V= number of valance electrons of central atom

X = number of monovalent atoms

C= charge on cation

A = charge on anion
 

Test: Hybridisation - Question 8

Specify the coordination geometry and hybridisation of N and B atoms in a 1:1 complex of BF3 and NH3.




Detailed Solution for Test: Hybridisation - Question 8


After 1 :1 complex is formed in which NH3 is electron donor and BF3 is electron acceptor

Each atom (N, B) both are sp3 hybridised. Since, lone pairs of N-atoms are also involved in bonding, hence,both are sp3 hybridized and tetrahedral.

Test: Hybridisation - Question 9

If there are five electron pairs in outer shell, then structure and bond angle as predicted by Sidgwick-Powell theory is

Detailed Solution for Test: Hybridisation - Question 9

Three electron pairs are inclined at 120°.
Two electron-pairs are at 90° with the first three.

Structure is trigonal bipyramidal.

Test: Hybridisation - Question 10

Hybridisation of Acetylene is

Detailed Solution for Test: Hybridisation - Question 10

Since acetylene is made up of triple bond. So the hybridization of carbon in acetylene is sp.

Test: Hybridisation - Question 11

Based on VSEPR model, AX4E representation is for

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Test: Hybridisation - Question 12

Direction (Q. Nos. 12 and 14) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.

Q. 

Statement I : BF3 molecule is planar but NF3 is pyramidal.

Statement II : N atom is smaller than B.

Detailed Solution for Test: Hybridisation - Question 12


(Ip-bp) repulsion make it pyramidal.
Thus, Statement I and II both are correct but Statement II is not the correct explanation of Statement I.

Test: Hybridisation - Question 13

Statement I : B atom is sp2-hybridised in B2H6.
Statement II : There is no lone pair or unpaired electron in B2H6.

Detailed Solution for Test: Hybridisation - Question 13

(d) B2H6 has following types of bonding

Bridging H-atoms and B-atoms are electron deficient. Each B-atom is however, joined to four H-atoms thus sp3-hybridised.
There is no lone pair or unpaired electron.
Thus, Statement I is incorrect but Statement II is correct.

Test: Hybridisation - Question 14

Statement I : BCI3 exists as monomer while AICI3 exists as dimer.

Statement II : Boron atom is too small to form the stable 4-membered ring system

Detailed Solution for Test: Hybridisation - Question 14


Size of B is too small to form stable 4-membered ring system. (Also delocalised n-bonding is disturbed). Hence, dimerofBCI3is not formed.
Dimer AI2CI6 is stable due to stable ring

Thus, Statements I and II both are correct and Statement II is the correct explanation of Statement 

*Multiple options can be correct
Test: Hybridisation - Question 15

Direction (Q. Nos. 15-17) This section contains 3 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or  MORE THANT ONE  is correct.

Q. 
Select the correct order of bond lengths of the specified bonds.

Detailed Solution for Test: Hybridisation - Question 15


Due to resonance (C— O) bond length is larger than but smaller than (C— O).



*Multiple options can be correct
Test: Hybridisation - Question 16

NH4NO3 and NH4NO2 differ is

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(a) Both N in anion are sp2-hybridised.

(c) Decomposition products are different.

*Multiple options can be correct
Test: Hybridisation - Question 17

Select the correct statement(s) about ethene.

Detailed Solution for Test: Hybridisation - Question 17


(C— H ) bond length < (C— H ) bond length
Nodal, plane of π-bond is located in molecular plane.

Test: Hybridisation - Question 18

Direction (Q. Nos. 18-19) This section contains  a paragraph, each describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d).

In certain polar solvents, PCI5 undergoes an ionisation reaction as shown

 

Q. Select the incorrect statement (s).

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Test: Hybridisation - Question 19

In certain polar solvents, PCI5 undergoes an ionisation reaction as shown

 

Q. Number of lone pairs on P in I, il and III are

Detailed Solution for Test: Hybridisation - Question 19

 Lone pair on each phosphorus atom = 0

Test: Hybridisation - Question 20

Direction (Q. Nos. 20-21) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or  MORE THANT ONE  is correct.

Q. Match the following compounds given in Column I with their geometry given is Column II
 


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Test: Hybridisation - Question 21

Match the processes in Column I with type of changes in hybridisation in Column II.
 

Detailed Solution for Test: Hybridisation - Question 21






 

*Answer can only contain numeric values
Test: Hybridisation - Question 22

Direction (Q. Nos. 22) This section contains 1 questions. when worked out will result in an integer from 0 to 9 (both inclusive)

Q. Most acidic H in the following compound is attached to C-atoms denoted by at S-N. ( ) ...


Detailed Solution for Test: Hybridisation - Question 22

Electronegativity of
sp3 < sp2 < sp-hybridised C-atom
(Greater the s-character, greater the electronegativity).
(C— H) electron pair is taken out by most (EN) atom most easily, making it is acidic.

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