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Trending Questions
Q.
Calculate the bond order of
Q. which bond has greater energy sigma or pi bond?
Q. why the bond order of CO+ is 3.5 and not 2.5 ?
Q. Paramagnetism is exhibited by the molecules which:
- are not attracted by magnetic field
- contains only paired electrons
- carries a positive charge
- contains unpaired electrons
Q.
In PCl5, the three P- Cl bonds in plane are called as
Sigma bonds
Equatorial bonds
Semi - planar bonds
Axial bonds
Q. 23. The wave length of the third line of the Balmer series for a hydrogen atom is
Q. Bond order depends on the number of bonding and anti-bonding orbitals. The bond order can never have:
- A negative value
- An integral value
- Assume any positive, or an integral, or a fractional value including zero
- a non zero value
Q. The bond dissociation energies of X2, Y2 and XY are in the ratio of 1:0.5:1 . ∆H for the formation of XY is -200KJ mol-1 . The bond dissociation energy of X2 will be
Q. More than One Answer Type
एक से अधिक उत्तर प्रकार के प्रश्न
Analyse the following graphs for a given orbital and select the correct statement(s).
दिए गए कक्षक के लिए निम्नलिखित ग्राफ का विश्लेषण कीजिए तथा सही कथन/कथनों का चयन कीजिए।
एक से अधिक उत्तर प्रकार के प्रश्न
Analyse the following graphs for a given orbital and select the correct statement(s).
दिए गए कक्षक के लिए निम्नलिखित ग्राफ का विश्लेषण कीजिए तथा सही कथन/कथनों का चयन कीजिए।
- ψ is maximum at nucleus but P(r) is zero because of negligible volume component
नाभिक पर ψ अधिकतम होता है लेकिन नगण्य आयतन घटक के कारण P(r) शून्य होता है - Given orbital has 1 radial node
दिए गए कक्षक में 1 त्रिज्य नोड है - Given orbital has zero angular nodes
दिए गए कक्षक में कोणीय नोड शून्य है - Given orbital is 2s
दिया गया कक्षक 2s है
Q. Match the following:Column-IColumn -II(a) O2−2(p)Bond order = 0.5(b) CO(q)Bond order = 3(c)NO+(r)Bond order = 1(d)He+2(s)Paramagnetic(t) Diamagnetic
- (a−r, t)
(b−q, t)
(c−q, t)
(d−p, s) - (a−q, t)
(b−r, t)
(c−r, t)
(d−p, s) - (a−r, s)
(b−q, s)
(c−r, s)
(d−r, t) - (a−q, s)
(b−p, s)
(c−p, t)
(d−r, t)
Q. Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n=1, l=0;
(b) n = 3; l=1;
(c) n = 4; l =2;
(d) n=4; l=3.
(a) n=1, l=0;
(b) n = 3; l=1;
(c) n = 4; l =2;
(d) n=4; l=3.
Q.
The number of π and σ bonds present in H2C=CHCH=CHCH2 are:
2, 12
11, 2
2, 11
12, 2
Q.
A. H2 1. 2
B. N2 2. 1
C. O2 3. 3
D.H2− 4. 0.5
The correct match of bond order is
A - 2, B - 3, C - 4, D - 3
A - 2, B - 1, C - 3, D - 4
A - 1, B - 2, C - 3, D - 4
A - 2, B - 3, C - 1, D - 4
Q. among F2, Cl2, O2 and N2 which has least bond dissociatuon enthalpy
Q. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals :
Q. In calculating the number of spectral lines when to apply these two formulas-
1. (n2-n1)(n2-n1+1)/2
2. (n2-n1)
Q. The radial wave equation for hydrogen atom is
Ψ=116√π(1a0)3/2[(x−1)(x2−8x+12)]e−x/2
Ψ=116√π(1a0)3/2[(x−1)(x2−8x+12)]e−x/2
where, x=2r/a0; a0= radius of first Bohr orbit.
The minimum and maximum position of radial nodes from nucleus are:
The minimum and maximum position of radial nodes from nucleus are:
- a0, 3a0
- a02, 3a0
- a02, a0
- a02, 4a0
Q.
Use molecular orbital theory to explain why the Be2 molecule does not exist.