MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Games
NEET
NEET Chemistry MCQ
Molecular Bonding Mcq Neet Chemistry
Quiz 6
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
The boiling point of water is 100°C where as that of hydrogen sulphide (H2S is -42°C. this can be attributed to
0%
a) a larger bond angle in water than hydrogen sulphide
0%
b) smaller size of oxygen atom as compared to sulphur
0%
c) larger ionization energy of oxygen than sulphur
0%
d) larger tendency of water to form hydrogen bonds than Hydrogen sulphide
Explanation
Since Oxygen is more electronegative than Sulphur thus strong hydrogen bonding in water Answer: (d)
Q.2
Which of the following compound has the least tendency to form hydrogen bonds between molecules
0%
a)NH3
0%
b) NH2OH
0%
c)HF
0%
d)CH3F
Explanation
Hydrogen is not bonded with F in option 'd'Answer: (d)
Q.3
In ice, the length of H-bonds
0%
a)is less than that of covalent bonds
0%
b) is greater than that of covalent bonds
0%
c)is same as that of covalent bonds
0%
d)can be less, greater or same as that of covalent bonds
Explanation
Answer: (b)
Q.4
The H-bonds in solid HF can be best represented as
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (c)
Q.5
Highest viscosity is exhibited by
0%
a) Glycerol
0%
b) Ethyl glycol
0%
c) Ethanol
0%
d) Water
Explanation
Due to the presence of -OH groups glycerol undergoes extensive intermolecular H bonding and hence it has highest viscosity Answer: (a)
Q.6
The crystal lattice of ice is mostly formed by
0%
a) ionic forces
0%
b) covalent bonds
0%
c) intermolecular as H-bonds
0%
d) covalent as well as H-bonds
Explanation
Answer:(d)
Q.7
on Q158) Out of the two compounds shows below, the vapour pressure of B at a particular temperature is expected to be
0%
a) higher than that of A
0%
b) lower than that of A
0%
c) same as that of A
0%
d) can higher or lower depending upon the size of the vessel
Explanation
Due to intramolecular H bonding in o-nitrophenol it exists as a monomer while due to intermolecular H bonding in p-nitrophenol it exists as an associated molecule. As a result, vapour pressure of o-nitrophenol is higher than that of p-nitrophenol Answer: (a)
Q.8
The correct order of the strength of H-bonds is
0%
a)H...F > H...O > H...N
0%
b) H...N > H...O>H...F
0%
c)H...O > H...N > H....F
0%
d)H...F > H...N > H...O
Explanation
Strength depends on electro negativity Answer:(a)
Q.9
There is no hydrogen bonding in
0%
a) Acetic acid
0%
b) Ammonia
0%
c) Ethyl alcohol
0%
d) Diethyl ether
Explanation
Diethyl ether does not contain a H-atom attached to O and hence does not undergo H-bonding Answer: (d)
Q.10
Out of CHCl3, CH4 and SF4 the molecule having regular geometry are
0%
a)CHCl3only
0%
b) CHCl3o and SF4
0%
c)CH4
0%
d)CH4 and SF4
Explanation
Due to absence of lone pairs of electrons on carbon, CH4 has regular tetrahedral geometry. Although CHCL3 also does not have loan pairs of electrons on carbon but due to the presence of three polar C-Cl bonds and one almost non polar C-H bond, its geometry is little distorted from regular shape. While SF4 has one lone pair and four bond pairs. therefore does not have regular geometry. Thus potion 'c' is correctAnswer: (c)
Q.11
Which among the following molecules is not flat
0%
a) C6 H6
0%
b) C2H4
0%
c)SO3
0%
d)C2H6
Explanation
C2H6 is sp3 hybridized while other options are sp2 hybridizedAnswer: (d)
Q.12
The compound in which C uses its sp3 hybrid orbitals for bond formation is ... [ IIT 1989]
0%
a) HCOOH
0%
b) (H2N)2CO
0%
c)HCHO
0%
d)CH3CHO
Explanation
CH 3 in CH3CHO have sp3 hybridizationAnswer: (d)
Q.13
The octet rule is not valid for the molecule .. [ IIT 1979]
0%
a)CO2
0%
b) H2O
0%
c)O2
0%
d)CO
Explanation
The central atom carbon has four bonds which means 4×2 = 8e- Hence, in CO2 cotect is followed There are two bond pair and 2 lone pair and thus the oxygen atom has a total of eight e-. Thus its octet rule is satisfied. O2: In oxygen there exist two bond between each O atom and they have two lone pairs each, so there are four pair of electrons, which means 4×2=8e- on each atom. Thus, its octet rule is satisfied. In CO molecule, the carbon atom contains only 6 valence electrons, where as 8e- are require to fulfill its octet. Hence, CO molecule does not have an octet Answer: (d)
Q.14
The structure of IF5 can be best described as
0%
a)
0%
b)
0%
c)
0%
d)non of these
Explanation
Total number of valence electrons in IF5 are=7 + 5 ×7=42 Now 42 ÷8=5(Q1 + 2(R1) R1 ÷ 2=2 ÷ 2=1(Q2) + 0 (R2) Now Q1+ Q2+R2=5+1+0=6 ∴ Hybridization of I=sp3d2 Since If5 has only five F atoms, therefore, sixth orbital will be occupied by a lone pair. So option 'a' is correctAnswer: (a)
Q.15
In N atom in NH4+ ion involves the hybridization
0%
a) sp
0%
b) sp2
0%
c) sp3
0%
d) sp3d
Explanation
Answer: (c)
Q.16
Sulphure in H2SO4 has hybridization
0%
a)sp3d3
0%
b) sp3d2
0%
c)sp3d
0%
d)sp3
Explanation
Number of valence electrons in H2SO4 are =2 ×1 + 6 + 4×6=32 Now 32÷8=4 (Q)+ 0(R)=4sp3 hybridizationAnswer: (d)
Q.17
in one of the following molecule, the state of hybridization of the central atom is not the same as in the others
0%
a) B in BF3
0%
b) O in H3O+
0%
c)N in NH3
0%
d)P in PCl3
Explanation
Total valance electrons in BF3 is >=3 +3 ×7=24 Now 24 ÷ 8=3(Q) + 0(R) Thus Q+ R=3, sp>3 hybridization In H3O+ free electrons are=3 ×1 + + -1=8 Now 8 ÷2=4(Q) + 0(R)=4 Thus sp>3 hybridization in NH3 number of valence electrons=5 + 3 ×1=8 ∴ Hybridization is sp3 Total number of valence electrons in PCl3 =5+ 3 ×7=26 NOw 26÷8=3(q1 + 2 (R1) and R1 ÷ 2=1(Q2 + 0 (R2) By adding all quotients and final remainder=3+1+0=4 ∴ Hybridization in of P in PCl3 is sp3Answer: (a)
Q.18
The hybrid state of C atom in charcoal is
0%
a) sp3
0%
b) sp2
0%
c)sp
0%
d)no specific state
Explanation
Answer:(d)
Q.19
Which of the following does not contain any coordinate bond
0%
a) H3O+
0%
b) BF4-
0%
c) HF2-
0%
d) NH4+
Explanation
In option a: Coordnate bond is formed between lone pair of oxygen atom and empty s-orbital of H atom. Option b: Boron atom has empty 2p orbital after formation of BF3, So if can accept lone pair of fluorine atom and form BF4- In Option c: There is no coordinate bonding but very strong hydrogen bonding dye to high electronegativity of flourine. In Option d: Nitrogen has lone pair left after forming three covalent bonds with hydrogen, it forms coordinate bond by sharing that lone pair with H-atom Only HF2- has H bonding [ F-H ...F]- rest all the molecules have coordinate bonds Answer: (c)
Q.20
Which of the following does not have linear structure
0%
a)HgCl2
0%
b) CS2
0%
c)SnCl2
0%
d)C2H2
Explanation
SnCl2 is bent moleculeAnswer: (c)
Q.21
The hybridization of the central atom is ICl2+ is
0%
a) dsp2
0%
b) sp
0%
c)sp2
0%
d)sp3
Explanation
Number of valence electrons in ICl2+ are=7 + 2 ×7 -1=20 Now 20 ÷ 8=2(Q1) + 4(R1)R1 ÷ 2=2(Q2 + 0 (R2Q1+Q2+R2=2+2+0=4sp3 hybridizationAnswer: (d)
Q.22
The atomic number of Sn isThe shape of gaseous SnCl2 molecule is
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Sn have two bond pairs and one loan pair hence structure 'b' is correct Answer:(b)
Q.23
Which carbon is more electronegative
0%
a) sp3 hybridized carbon
0%
b) sp hybridized carbon
0%
c) sp2 hybridized carbon
0%
d) the electron attracting power of C is always same irrespective of its hybrid state
Explanation
Higher the s- character of the hybrid orbital, more is its electro negativity. Thus sp-hybrid carbon is most electronegative Answer: (b)
Q.24
The maximum number of covalent bonds by which the two atoms can be bonded to each other is
0%
a)Four
0%
b) Two
0%
c)Three
0%
d)no fixed number
Explanation
The hybridization which leads to the formation of maximum bonds between two atoms is sp. therefore, at the maximum three bonds can be formed between two atoms example C-C ≡ N ; C-C = O Orbital geometry around carbons can not allow them to overlap in such a way that they form 4 bonds Answer: (c)
Q.25
The hybridization of phosphorus in POCl3 is the same as in
0%
a) P in PCl3
0%
b) S in SF4
0%
c)Cl in ClF3
0%
d)B in B Cl3
Explanation
Valence electrons in POCl3 is 5 +6+3 ×7=32 Now 32÷8=4(Q) + 0(R)since Q+ R=4+0=4 sp3 hybridization Total number of valence electrons in PCl3 =5+ 3 ×7=26 NOw 26÷8=3(q1 + 2 (R1) and R1 ÷ 2=1(Q2 + 0 (R2) By adding all quotients and final remainder=3+1+0=4 ∴ Hybridization in of P in PCl3 is sp3Answer: (a)
Q.26
he hybrid state of C atom in C2H2 is same as that of carbon in
0%
a) C2H6
0%
b) CO2
0%
c)Benzene
0%
d)C (diamond)
Explanation
In both C2H2 and CO2 c is sp2 hybridized Answer:(b)
Q.27
Incorrect order of decreasing boiling point is
0%
a) HF > HI > HB > HCl
0%
b) H2O > H2Te > H2Se > H2S
0%
c) Br2 > Cl2 > F2
0%
d) CH4 > GeH4 > SiH4
Explanation
Ch4 do not form hydrogen bond hence boiling point should be lowest in the series Answer: (d)
Q.28
Which of the following statement about repulsion between bond pairs(bp) and loan pairs(lp) is correct
0%
a)lp-lp > lp-bp > bp-bp
0%
b) lp-bp > lp-lp > bp-bp
0%
c)bp-bp > lp-bp > lp-lp
0%
d)Any of the three depending upon the type of molecule
Explanation
Answer: (a)
Q.29
Which of the following statement is incorrect for PCl5
0%
a) Its all P-Cl bond lengths are equal
0%
b) It involves sp3d hybridization
0%
c)It has an irregular geometry
0%
d)Its shape is trigonal bipyramidal
Explanation
Answer:(a)
Q.30
In which of the following molecules, the central atom does not use sp3 hybrid orbitals in its bonding
0%
a) NH2-
0%
b) BeF3-
0%
c)SO2Cl2
0%
d)SO42-
Explanation
Total electrons in BeF3- =2 +3 ×7 +1=24Now 24÷8=3(Q)+ 0(R)=3 Thus Be in BeF3- is sp2 hybridized All other molecules, the hybridization of central atom is sp3Answer: (b)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Support mcqgeeks.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page