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CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 15 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Chemical Bonding And Molecular Structure
Quiz 15
VSEPR theory suggest a small departure from $$ 109^0 28'$$ for $$ NH_3 $$ molecule. The H-N-H bond angle has been found to be $$ 107^0 $$ what kind of hybrid orbital will be occupied by the non -bonding pair of electron?
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$$ sp^3 $$
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$$ sp^{3.4} $$
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$$ sp^{2.13} $$
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$$ sp^2 $$
Which of the following has -O-O- linkage ?
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$$ H_2S_2O_6 $$
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$$ H_2S_2O_8 $$
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$$ H_2S_2O_3 $$
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$$ H_2S_4O_6 $$
Which species has the maximum number of lone pair of electrons on the central atom ?
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$$ [ ClO^-_3] $$
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$$ XeF_4 $$
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$$ SF_4 $$
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$$ [ I^-_3] $$
Which of the following sets of molecule(s) is/are having a linear shape but different hybridization?
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$$I^-_3$$ and $$CO_2$$
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$$HgCl_2$$ and $$ICl^-_2$$
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$$XeF_2$$ and $$C_2H_2$$
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$$XeF_2$$ and $$OCl^-$$
Explanation
A) $$I_{3}^{-} $$ and $$ CO_{2}$$ are linear in shape.
$$I_{3}^{-} $$ undergoes $$sp^{3} $$ hybridization and $$ CO_{2}$$ undergoes $$ sp $$ hybridization.
B) $$HgCl_{2} $$ and $$ICl_{2}^{-} $$ are linear in shape.
$$HgCl_{2} $$ undergoes $$ sp$$ hybridization and $$ICl_{2}^{-} $$ undergoes $$ sp^{3}d $$ hybridization .
C) $$XeF_{2} $$ and $$ C_{2}H_{2} $$ are linear in shape.
$$ XeF_{2} $$ undergoes $$sp^{3}d^{2} $$ and $$C_{2}H_{2} $$ undergoes $$ sp $$ hybridization.
D) $$XeF_{2} $$ and $$ OCl^{-} $$ are linear in shape.
$$ XeF_{2} $$ undergoes $$sp^{3}d^{2} $$ and $$ OCl^{-} $$ undergoes $$ sp^{3} $$ hybridization .
Hence, options A, B, C, and D all are correct.
In $$ OF_2 $$ number of bond pairs and lone pairs of electrons are:
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2, 6
0%
2, 8
0%
2, 10
0%
1, 9
Explanation
In $$OF_2$$ molecule, the total number of bond pairs and lone pairs of electrons present is 2 and 8.
There are two bond pairs of electrons which form two $$O-F$$ pairs. $$O$$ atom has two lone pairs of an electron. Each $$F$$ atom has 3 lone pairs of electrons. Thus in total, there are 8 lone pairs of electrons.
Hence the correct answer is option B.
A diatomic molecule has a dipole moment of $$1.2 D$$ if its bond distance is $$1.0 \mathring{A}$$. What fraction of an electronic charge 'e' (i.e. charge on an electron) exists on each atom?
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$$20\%$$ of e
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$$21\%$$ of e
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$$19\%$$ of e
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$$25\%$$ of e
Which of the following molecule/species is having minimum number of lone pair on its central atom?
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$$BrF_3$$
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$${BrF_4}^-$$
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$${XeF_5}^+$$
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$${I_3}^-$$
Which of the following statement is/are correct ?
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All carbon to carbon bonds contains a sigma bond and one or more $$\pi$$-bonds
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All carbon to carbon bonds are sigma bonds
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All oxygen to hydrogen bonds are hydrogen bonds
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All carbon to hydrogen bonds are sigma bonds
In $$ NO_3^- $$ ion, number of bond pairs and lone pairs of electrons on nitrogen atom are -
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2, 2
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3, 1
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1, 3
0%
4, 0
Explanation
In $$NO_3^-$$ ion, due to the presence of one negative charge, number of valence electrons $$=5+1=6$$
One $$O-$$atom forms two bonds
and two $$O-$$atom are shared with two electrons of $$N-$$atom
Thus, $$3$$ $$O-$$atoms are shared with $$8$$ electrons of $$N-$$atom.
Number of bond pairs (or shared pairs) $$=4$$
Number of lone pairs $$=0$$
Hence correct answer is option D.
In $$OF_2$$, the number of bond pairs and lone pairs of electrons is, respectively:
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$$2, 6$$
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$$2, 8$$
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$$2, 10$$
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$$2, 9$$
The formula charges on three oxygen atoms of ozone molecule are respectively:
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+1, 0 , + 1
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0, 0 , 0
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+1, 0 , -1
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-1 , +1 , -1
Explanation
According to the electron-dot structure, there are 12 number of bonding electrons and 6 number of non-bonding electrons.
Now we have to calculate the formal charges of each oxygen atom.
Formal charge $$=$$ Valence electrons $$-$$ Non-bonding electrons $$-\dfrac{Bonding\,\,electrons}{2}$$
Formal charge on $$O_1=6-2-\dfrac 62=+1$$
Formal charge on $$O_2=6-4-\dfrac 42=0$$
Formal charge on $$O_3=6-6-\dfrac 22=-1$$
Hence the correct answer is option C.
Covalency is favoured in the following cases:
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a smaller cation
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a larger anion
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large charges on cation and anion
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a cation without noble gas configuration
$$C-C$$ bond length is maximum in:
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diamond
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graphite
0%
naphthalene
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fullerene
Identify the correct sequence of increasing number of $$\pi-$$ bonds in structures of the following molecules.
(i) $$H_{2}S_{2}O_{6}$$
(ii) $$H_{2}SO_{3}$$
(iii) $$H_{2}S_{2}O_{5}$$
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$$I, II, III$$
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$$II, III, I$$
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$$II, I, III$$
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$$I, III, II$$
The number of $$P-O-P$$ bonds in cyclic metaphosphoric acid is:
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zero
0%
two
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three
0%
four
Compound $$X$$ is anhydride of sulphuric acid. The number of sigma bonds and $$\pi$$- bonds present in $$X$$ are respectively:
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$$6$$ and $$6.95$$
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$$5$$ and $$5.95$$
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$$3$$ and $$3$$
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$$4$$ and $$5.92$$
The bond length of the species $$O_2, O_2^+,and\ O_2^-$$ are in the order of:
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$$O_2^+ >O_2 >O_2^-$$
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$$O_2^+ >O_2^- >O_2$$
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$$O_2 >O_2 ^+>O_2^-$$
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$$O_2^->O_2 >O_2^+$$
Explanation
As bond order increases bond length decreases the bond order of species are:
$$\dfrac {Number \, of \, bonding \, electron- Number \, of a.b. \, electron}{2}$$
For $$O_2=\dfrac {10-6}{2}=2$$
$$O_2^+=\dfrac {10-5}{2}=2.5$$
$$O_2^-=\dfrac {10-7}{2}=1.5$$
so bond order $$O_2^+ >O_2 >O_2^-$$ and bond length are $$O_2^- >O_2 >O_2^+$$
Option D is correct.
Which species has the maximum number of lone pairs of electrons on the central atom?
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$$ClO_3^-$$
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$$XeO_4$$
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$$I_3^-$$
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$$SF_4$$
The bond strength increase:
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with increasing bond order
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with increasing extent of overlapping of orbitals
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with decreasing in bond length
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from a single bond to triple bond
In which pair(s) is/are the stronger bond found in the first species?
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$$ {O_{2}}^{-}, O_{2} $$
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$$ N_{2}, {N_{2}}^{+}$$
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$${NO}^{+}, {NO}^{-}$$
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$$ CO,{O_{2}}^{2+}$$
Number of sigma bonds in $$P_4O_{10}$$ is:
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6
0%
7
0%
17
0%
16
Explanation
Number of sigma bonds in $$P_4O_{10}$$ is $$(d)$$ $$16$$.
In which case the bond length is minimum between carbon and nitrogen?
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$$CH_3NH_2$$
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$$C_6H_5CH=NOH$$
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$$CH_3CONH_2$$
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$$CH_3CN$$
Explanation
In $$CH_3CN$$ bond order between C and N is 3 so its bond length is minimum.
There is triple bond between C and N in $$CH_3CN$$
Option D is correct.
How many bonding electron pairs are there in white phosphorous?
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6
0%
12
0%
4
0%
8
The number of $$P-O-P$$ bonds in cyclic meta phosphoric acid is:
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one
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two
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three
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four
Explanation
The number of $$P-O-P$$ bond in cyclic metaphosphoric acid is three.
In benzene the total number of $$\sigma$$ bonds is
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$$3$$
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$$6$$
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$$9$$
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$$12$$
Explanation
$$C-C\ \sigma$$ bonds $$=6$$
$$C-H\ \sigma $$ bonds $$=6$$
__________
Total sigma bond $$=12$$
Which of the following statements are correct about $$CO_{3}^{2-}$$?
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The hybridisation of central atom is $$sp^{3}$$.
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Its resonance structure has one $$C-O$$ single bond and two $$C=O$$ double bonds.
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The average formal charge on each oxygen atom is $$0.67$$ units.
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All $$C-O$$ bond lengths are equal.
Explanation
The hybridisation of central atom is $$sp^2$$
Since a net charge of $${-}2$$ is shared over 3 $$Oxygen$$ atoms, hence formal charge = $$0.67$$ units
In resonating structures bonds are not fixed and therefore, all bond lengths are equal.
hence options $$(C)$$ and $$(D)$$ are correct.
The bond in $$K_{4}[Fe(CN)_{6}]$$ are
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all ionic
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all covalent
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ionic and covalent
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ionic, covalent and co-ordinate covalent
Explanation
The bond in $$K_{4}[Fe(CN)_{6}]$$ are ionic, covalent and co-ordinate covalent.
Ionic bond between cation and anion, covalent bond between C and N atom in the ligand and coordinate bond between ligand and central metal atom.
Which of the following has no net dipole moment?
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0%
0%
0%
Number of bonds in benzene
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$$6\sigma$$ and $$3\pi$$
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$$12\sigma$$ and $$3\pi$$
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$$3\sigma$$ and $$12\pi$$
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$$6\sigma$$ and $$6\pi$$
Explanation
The number of bonds in benzene is $$12\sigma$$ and $$3\pi$$ bonds.
Number of $$\sigma$$ bonds in
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$$6$$
0%
$$15$$
0%
$$10$$
0%
$$12$$
Explanation
There are $$15\ \sigma$$ and $$3\ \pi$$ bonds in the given molecule. Each double bond contain $$1\ \sigma$$ and $$1\ \pi$$ and single bond form is always a sigma bond.
Which of the following statements is wrong?
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Single $$N-N$$ bond is stronger than the single $$P-P$$ bond.
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$$PH_{3}$$ can act as a ligand in the formation of coordination compound with transition elements.
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$$NO_{2}$$ is paramagnetic in nature.
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Covalency of nitrogen in $$N_{2}O_{5}$$ is four.
Explanation
N-N sigma bond (single bod) is weaker than the P-P sigma bond (single bond) due to the small bond length between the nitrogen atoms. The lone pair of electrons of both the atoms nitrogen repel each other making the single bond between N−N weaker than P-P sigma bond. In P−P bond the repulsion between sigma bond and lone pair electrons is less as the size of phosphorous is large.
So statement A is wrong.
Hence option A is correct.
In a compound the number of sigma and pi bonds respectively are :
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19, 11
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19, 5
0%
13, 11
0%
7, 3
Explanation
It is known that all single bond is sigma bond while multiple bond is made of 1 sigma
bond and rest pi
bond.
here
between N and C have 1 sigma and 2 pi bond
between c and O have 1 sigma and 2 pi bond
M and C have 3 sigma bonds
now on counting all nonds
we get sigma bond =19
pi bond =11
Which of the following compounds have the same no. of lone pairs with their central atom?
(I) $$XeF_5^-$$ (II) $$BrF_3$$ (III) $$XeF_2$$ (IV) $$H_3S^+$$ (V) Triple Methylene
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IV and V
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I and III
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I and II
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II, IV and V
Explanation
$$XeF_5^-$$ : 2 lone pairs : pentagonal planar
$$BrF_3$$ : 2 lone pairs : Bent-T-shape
$$XeF_2$$ : 3 lone pairs : Linear
$$H_3S^+$$ : 1 lone pairs : Pyramidal Linear
Triple Methylene : 1 lone pair : V-shape
Correct set of species with zero dipole moment is :
(i) $$CO_{2}$$ (ii) $$COCl_{2}$$ (iii)$$ CH_{2}Cl_{2}$$ (iv) $$BCl_{3}$$
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(i) and (iv)
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(ii) and (iv)
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(iii) and (iv)
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(i), (iii) and (iv)
Which of the following combination of orbitals does not form covalent bond (x-axis is inter nuclear axis) :
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$$s + p_{y}$$
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$$p_{y} + p_{y}$$
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$$d_{yz} + d_{yz}$$
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$$d_{xy} + d_{xy}$$
Covalency favoured in the following case :
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smaller cation
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larger anion
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large charge on cation and anions
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all of these
Explanation
According to Fajan"s rule-
The compound with big size of anion and small size size of cation is more covalent in nature
The anion or cation with higher charges can show covalent character
The anion or cation with low charge can show ionic charecter
Consider the following compounds and select the incorrect statement from the following :
$$NH_{3}, PH_{3}, H_{2}S, SO_{2}, SO_{3}, BF_{3}, PCl_{3}, IF_{7}, P_{4}, H_{2}$$
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Six molecules out of given compound involves hybridisation
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Three molecules are hypervalent compounds
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Six molecules out of above compounds are non-planar in structure
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Two molecules out of given compounds involves $$(d \pi- p \pi)$$ bonding as well as also involves $$(p \pi - p \pi)$$ bonding
The incorrect statement(s) regarding $$\overset .{C} X_3$$ species is :
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If electronegativity of surrounding element $$'X'$$ is less than $$2.5$$, then central carbon atom used almost $$33 \%$$ s-character in their hybrid bonding orbitals.
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If electronegativity of surrounding element $$'X'$$ is greater than $$2.5$$, then central carbon atom used almost $$25 \%$$ s-character in their hybrid bonding orbitals.
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If $$'X'$$ is $$'F'$$, then species should be polar and pyramidal
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If $$'X'$$ is $$H$$, then species should be polar and planar
Number of lone pairs present on central atom of $$I_{3}^{-}$$ are:
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$$9$$
0%
$$1$$
0%
$$6$$
0%
$$3$$
Explanation
$${I_{3}}^{-}$$ has a linear structure. The number of lone pairs on central atom is $$3$$.
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