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Test: Chemical Bonding- 1 - Chemistry MCQ


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20 Questions MCQ Test - Test: Chemical Bonding- 1

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Test: Chemical Bonding- 1 - Question 1

The geometry of H2​S and its dipole moment are :

Detailed Solution for Test: Chemical Bonding- 1 - Question 1
  • H2​S molecule has an angular geometry because it has two lone pairs of electrons that make the molecule bend. The dipole moment is not zero.
  • This is because sulfur is more electronegative than hydrogen, making the molecule slightly polar.

Test: Chemical Bonding- 1 - Question 2

Which one of the following molecules is planar?

Detailed Solution for Test: Chemical Bonding- 1 - Question 2
  • NF3​,NCl3​,PH3​ - They all have trigonal pyramidal geometry.
  • They all have a lone pair of electrons on their central atom.
  • Whereas BH3​ does not have a lonee pair, and is planar.

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Test: Chemical Bonding- 1 - Question 3

Which of the following is the electron-deficient molecule?

Detailed Solution for Test: Chemical Bonding- 1 - Question 3
  • B2His electrons deficient molecule because boron atom has thre half-flled orbits in excited state.
  • In it two electrons of a B-H bond are involved in formation of three centre bond, these bonds are represented as dotted lines.
  • The structure of B2H6 is represented as follows.

Test: Chemical Bonding- 1 - Question 4

If a bond is made up of a large number of organic compound, then the bond is termed as ______.

Detailed Solution for Test: Chemical Bonding- 1 - Question 4

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. Organic Compounds are made up through sharing electron pairs.

Test: Chemical Bonding- 1 - Question 5

The hybridization, lone pair of electrons, and shape of I3​ is:

Detailed Solution for Test: Chemical Bonding- 1 - Question 5

Hybridization: to find the no.of hybridized bonds, add the no. of lone pairs with the no.of neighbour atoms.
3l.p+2(iodine atoms)=5
So Hybridization of l3− = sp3d 
Note: These are the hyb. numbers that represent the type of hybridization.

2∼sp

3∼sp2

4∼sp3

5∼sp3d

6∼sp3d2

lone pairs - 3
Shape - Linear with a bend angle of 180°. The lone pairs are arranged in a trigonal bipyramidal arrangement.

image

Test: Chemical Bonding- 1 - Question 6

Which compound species has all the carbons, with same kind of hybridization?

Detailed Solution for Test: Chemical Bonding- 1 - Question 6

Ortho, Meta & Para Benzene are all the same. i.e benzene only so all the carbons are having the same kind of hybridization i.e spso, Option D is correct.

Test: Chemical Bonding- 1 - Question 7

Which among the following formation is not an example of Covalent bond?

Detailed Solution for Test: Chemical Bonding- 1 - Question 7

LiF (Lithium Fluoride) is an example of Ionic bond, as the formation takes place by transfer of electrons and not by sharing.

Test: Chemical Bonding- 1 - Question 8

Which type of bond will be formed by overlapping of dxz and dxz orbitals if the molecular axis is x-axis?

Detailed Solution for Test: Chemical Bonding- 1 - Question 8

Lateral overlap along the X-axis between two dzx orbitals leads to the formation of a Pi-bond.

Test: Chemical Bonding- 1 - Question 9

Examples of refractory materials include:

Detailed Solution for Test: Chemical Bonding- 1 - Question 9

Magnesium Oxide(MgO) is used to make linings for some furnaces. It is known as a refractory material - which just means that it is resistant to heat and retains its strength at very high temperatures.

Test: Chemical Bonding- 1 - Question 10

In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic? 

Detailed Solution for Test: Chemical Bonding- 1 - Question 10

Diamagnetism is a quantum mechanical effect that occurs in all materials; when it is the only contribution to the magnetism, the material is called diamagnetic.

Paramagnetism is a form of magnetism whereby some materials are weakly attracted by an externally applied magnetic field and form internal induced magnetic fields in the direction of the applied magnetic field.

So, only in the conversion of NO → NO+, the bond order has increased from 2.5 to 3 and the paramagnetic character has changed to diamagnetic.

Test: Chemical Bonding- 1 - Question 11

The hybridization state on BH3 is:

Detailed Solution for Test: Chemical Bonding- 1 - Question 11

In BH3

We know that,

Boron has three valence electron, so it is supposed to make 3 bond in a molecules with hybridization sp2 as only S and two p are used in hybridization because last p orbital vacant.

Test: Chemical Bonding- 1 - Question 12

The hybridization state in triangular bipyramidal CH5+ is:

Detailed Solution for Test: Chemical Bonding- 1 - Question 12
  • Carbonium ion methonium,  CH5+ is produced by the addition of  H+ to CH4. In CH5+ three hydrogen atoms are in a plane and one above and one below the plane. 
  • So its sp3d and the shape is trigonal bipyramid. 
  • But this compound is considered a CH3+ carbenium ion with a molecule of hydrogen interacting with the empty orbital in a
    3-center-2-electron bond.
  • As the two hydrogen atoms in H2 are in continuously exchange positions with the three hydrogen atoms in the CH3+ methonium CH5+ is considered as a fluxional molecule. Due to the existence of CH3+, this is sp2​ hybridised. 
Test: Chemical Bonding- 1 - Question 13

 Which among the following chemical bond were described by Kossel and Lewis?

Detailed Solution for Test: Chemical Bonding- 1 - Question 13

Both Ionic and Covalent bond arise from the tendency of atoms to attain stable configuration of electrons.

Test: Chemical Bonding- 1 - Question 14

The percentage of p-character in the orbitals forming P−P bonds in P4 is

Detailed Solution for Test: Chemical Bonding- 1 - Question 14

The percentage of p-character in forming P-P bonds in P4 molecule is: 75. In forming 
Pmolecule, P atom uses sp3 hybrid orbitals. Thus, the p-character in hybrid orbitals is 75%.

Test: Chemical Bonding- 1 - Question 15

Which one shows maximum hydrogen bonding?

Detailed Solution for Test: Chemical Bonding- 1 - Question 15

The correct option is : a H2O Explanation:H2O shows maximum H-bonding because each H2O molecule is linked to four H2O molecules through H-bonds.

Test: Chemical Bonding- 1 - Question 16

The hybridization state in ‘B’ when BF3 form adduct with ether is:

Detailed Solution for Test: Chemical Bonding- 1 - Question 16

When BF3 forms an adduct with ether, it has 4 bonds respectively bonded with Oxygen (of ether) nd 3 F atoms. Presence of 4 bonds to the central atom results in sp3 hybridization.

Test: Chemical Bonding- 1 - Question 17

In which of the following pairs, both the species have the same hybridisation?
(I) SF4,XeF4 (II) I−3,XeF2 (III) ICl+4,SiCl4 (IV) ClO−3,PO3−4

Detailed Solution for Test: Chemical Bonding- 1 - Question 17

Test: Chemical Bonding- 1 - Question 18

In allene (C3H4), the type(s) of hybridisation of the carbon atoms is (are):

Detailed Solution for Test: Chemical Bonding- 1 - Question 18

An allene is a compound in which one carbon atom has double bonds with each of its two adjacent carbon centres.

The central carbon atom of allene forms two sigma bonds and two pi bonds. The central carbon is sp-hybridized, and the two terminal carbon atoms are sp2-hybridized. The bond angle formed by the three carbon atoms is 180°, indicating linear geometry for the carbon atoms of allene.

Test: Chemical Bonding- 1 - Question 19

With respect to hyper valence theory, which will have multicenter bonding:

Detailed Solution for Test: Chemical Bonding- 1 - Question 19

Multicenter bonding is typically associated with compounds that exhibit bonding that cannot be explained by conventional two-center two-electron bonds. This often involves compounds with electron-deficient atoms or those that form hypervalent bonds.

Let's evaluate the options:

1. B2H6: Diborane (B2H6) is a classic example of a molecule with multicenter bonding. It has two bridging hydrogen atoms that are involved in three-center two-electron bonds, connecting two boron atoms. This is a well-known case of multicenter bonding.
2. CH6++: This is not a standard compound and seems hypothetical. Such a molecule would be highly unstable and could theoretically involve multicenter bonding, but it's not a commonly recognized species.
3. SF4: Sulfur tetrafluoride (SF4) is hypervalent and has 10 electrons around sulfur, but it does not involve multicenter bonding. The bonding in SF4 can be explained using standard two-center bonds.

 

Given that B2H6 clearly has multicenter bonding and the other options are either incorrect or ambiguous:

Correct Answer: 1. B2H6

Test: Chemical Bonding- 1 - Question 20

Which of the following are isoelectronic and isostructural NO3​, CO3−2​, ClO3​, SO3

Detailed Solution for Test: Chemical Bonding- 1 - Question 20

Correct option is A) NO3,CO3−2

Number of electrons in the compound is :

  • NO3​= 32e−
  • CO3-2 ​= 32e−
  • ClO3​= 42e−
  • SO3​= 41e−

Therefore, NO3​ and CO3-2​ are isoelectronic

The structures of both NO3​ and CO3-2

 is trigonal planar.

Therefore NO3− ​and CO3-2are isostructural also.

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