The compound with a maximum number of lone pairs of electrons on
the central atom among the following is:
1.
2.
3.
4.
In the structure of , to decide the plane in which C = S is present the following bond angle value are given
Axial FSF angle
Equatorial FSF angle
After deciding the plane of double bond, which of the following statements is correct?
(1) Two C-H bonds are in the same plane of axial S-F bonds.
(2) Two C-H bonds are in the same plane of equatorial S-F bonds.
(3) Total five atoms are in the same plane.
(4) Equatorial S-F bonds are perpendicular to plane of bond.
Which of the following are isoelectronic and isostructural ?
(1)
(2)
(3)
(4) All of these
The correct order of acidic strength is :
1.
2.
3.
4.
In which of the following molecules / ions are all the bond angles not equal?
(1)
(2)
(3)
(4)
What is the correct order from the weakest to the strongest carbon-oxygen bond for the following species?
(1)
(2)
(3)
(4)
According to Molecular Orbital Theory, which of the following statement is incorrect?
1. C2 has no unpaired electrons.
2. In molecule, both the bonds are bonds.
3. is paramagnetic but is diamagnetic.
4. In ion, there is one and two bonds.
Match the ionization processes listed in column – I with the changes observed as listed in
column-II. For this use the codes given below:
Column- I Column - II
(a) (p) Bond order increases and magnetic property is changed.
(b) (q) Bond order decreases and magnetic property is not changed.
(c) (r) Bond order increases and magnetic property is not changed.
(d) (s) Bond order decreases and magnetic property is changed.
Note : Here change in magnetic property refers to change from diamagnetic to paramagnetic
or paramagnetic to diamagnetic.
a b c d
1. s q p r
2. s p q r
3. r q s p
4. p s q r
A diatomic molecule has a dipole moment of 1.2 D. If its bond length is equal to 10 -10 m , then the fraction of an electronic charge on each atom will be:
1. 42%
2. 52%
3. 37%
4. 25%
Arrange the following molecules from most to least polar.
(1) III>V>II > IV=I
(2) II>IV>III>IV>I
(3) III>II>V>IV>I
(4) V>III>II>IV=I