Chemical Bonding and Molecular Structure Test -6

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Chemical Bonding and Molecular Structure Test -6

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Weekly Quiz Competition

Question 1
1 / -0

The state of hybridization of the central atom and the number of lone pairs over the central atom in POCl₃are

A
sp, 0

B
dsp², 1

C
sp³, 0

D
sp², 0

Solution

CN= 0.5[V+M-C+A]

For POCl₃ V=5 M=3 C =A=0 So CN=4

so hybridization is sp³. And there are 4 bps so 0 lps.
Question 2
1 / -0

Which one of the following conversions involve change in both hybridization and shape?

A
CH₄→ C₂H₆

B
BF₃→ BF₄⁻

C
NH₃→ NH₄⁺

D
H₂O → H₃O⁺

Solution

BF₃ has sp² hybridisation and its shape is triangular planar While BF₄^- has sp³ hybridisation and shape is tetrahedral.
Question 3
1 / -0

Based on VSEPR theory, the number of 90 degree F — Br — F angles in BrF₅is

A
2

B
8

C
4

D
0

Solution

Steric no. of BrF₅ is 6 so geometry is octahedral. It will have 5 bps and one lp which is present on axial position. Because of distortion caused by lp there is no 90^o F—Br—F bond angle in BrF₅ .
Question 4
1 / -0

The species having pyramidal shape is

A
SiO²⁻₃

B
SO₃

C
BrF₃

D
SF₂O

Solution

SF₂O has pyramidal shape as it is sp³ hybridised and it has 3bp and 1 lp.
Question 5
1 / -0

For a stable molecule the value of bond order must be

A
negative

B
zero

C
there is no relationship between stability and bond order

D
positive

Solution

Bond order ∝ Stability

Hence, for a stable molecule the value of bond order must be positive. When bond order is zero and negative the molecule will not exist.
Question 6
1 / -0

In acetylene molecule, between the carbon atoms there are

A
two sigma and one pi bonds

B
one sigma and two pi bonds

C
three pi bonds

D
three sigma bonds

Solution

acetylene is C₂H₂.

C atoms are bound with triple bond i.e. 1 sigma and 2 pie bonds.
Question 7
1 / -0

In PO³⁻₄ ion the formal charge on the oxygen atom of P–O bond is

A
+ 0.75

B
– 0.75

C
– 1

D
+ 1

Solution

Formal charge on the oxygen atom of P–O bond is – 1.

Formal charge = ( no. Of Valence electrons) - ( no. Of non bonded electrons + no. Of bonds).
Question 8
1 / -0

In which of the following substances will hydrogen bond be strongest?

A
HCl

B
HI

C
H₂O

D
H₂S

Solution

Out of all H₂O will form strongest hydrogen bonding due to high electronegativity of O.
Question 9
1 / -0

Which of the following unit conversion of dipole moment is correct?

A
$$1 D = 3.3356 Cm$$

B
$$1D= 3.3356\times10^{-30}C m$$

C
$$1 Cm = 3.3356 D$$

D
$$1 C m = 3.3356\times10^{-30} D$$

Solution

One Debye,
$$1D=10^{-18} esu\cdot cm$$ esu. electrostatic unit
On coverting it in SI units, $$1D=3\cdot 33\times 10^{-30}$$ coloumb metre.
Question 10
1 / -0

Which of the following does not contain any co-ordinate bond?

A
$$H_3O^+$$

B
$$BF_4^-$$

C
$$HF_2^-$$

D
$$NH_4^+$$

Solution

In option A, coordinate/dative bond is formed between lone pair of Oxygen atom and empty s orbital of H ion.
In option B, Boron atom has empty 2p orbital after formation of BF3, so it can accept lone pair of fluorine atom (F atom has 3 lone pairs) and form BF4-
In option C, there is no coordinate bonding but very strong Hydrogen bonding due to high electronegativity of Fluorine. (C) is the correct answer.
In option D, Nitrogen has one lone pair left after forming 3 covalent bonds with hydrogen, it forms coordinate bond by sharing that lone pair with H ion.
Question 11
1 / -0

Which of the following statements is not correct?

A
$$(s + p_y)$$ produces sp hybrid orbitals which are lying in the yz plane

B
$$(s + p_y)$$ produces sp hybrid orbitals which are lying the xz plane

C
$$(s + p_x + p_z)$$ produces sp2 hybrid orbitals which are lying in the xz plane

D
$$(s p_y)$$ produces sp hybrid orbitals which are lying along the y-axis

Solution

$$(S+P_y)$$
Produces SP hybrid orbital which is orlinked in y- direction i.e. its nodal plane its xz plane.
Question 12
1 / -0

The correct order regarding the electronegativity of hybrid orbitals of carbon is:

A
$$sp > sp^{2}< sp^{3}$$

B
$$sp > sp^{2} > sp^{3}$$

C
$$sp < sp^{2}> sp^{3}$$

D
$$sp < sp^{2}< sp^{3}$$

Solution

$$sp>s{ p }^{ 2 }>s{ p }^{ 3 }$$
$$s$$ orbitals hold electrons more tightly to nucleus than $$p$$ orbitals. That is the $$s$$ orbitals are effectively more electronegative.
In $$s{ p }^{ 2 }$$ carbon, the character of each orbital has $$33$$% of s character and $$25$$% s character in $$s{ p }^{ 3 }$$ carbon, whereas in $$sp$$, $$s$$ character has $$50$$% and therefore the electronegativity order goes this way
$$sp>s{ p }^{ 2 }>s{ p }^{ 3 }$$
Question 13
1 / -0

Directions For Questions

In the valence bond theory, hybridisation of orbitals is an integral part of bond formation. Hybridisation consists

ofmixing or linear combination of the "pure" atomic orbitals in such a way as to form new hybrid orbitals such
as $$sp, sp^{2} , sp^{3} ,sp^{3}d, sp^{3}d^{2}$$ etc.

...view full instructions

Which one of the following molecular geometries (i.e. shapes) is not possible for the $$sp^{3}d^{2}$$ hybridization?

A
Capped octahedral

B
Octahedral

C
Square planar

D
Square pyramidal

Solution
Question 14
1 / -0

The molecule in which xenon is surrounded by three lone pairs and two bond pairs is:

A
$$ \mathrm{X}\mathrm{e}\mathrm{F}_{2}$$

B
$$\mathrm{X}\mathrm{e}\mathrm{F}_{4}$$

C
$$\mathrm{X}\mathrm{e}\mathrm{F}_{6}$$

D
$$\mathrm{X}\mathrm{e}\mathrm{O}_{4}$$

Solution

The number of lone pair $$= 3$$ and bond pair $$= 2$$ in $$XeF_2$$.
Question 15
1 / -0

Select the INCORRECT statement :

A
$$N_2O$$ with sodium metal in liquid ammonia forms sodium azide and nitrogen gas is liberated.

B
Ammonia is oxidized to nitrogen by dilute solution of sodium hypochlorite in presence of glue.

C
Ammonium dichromate on heating decomposes to give nitrogen and a green coloured compound.

D
CaNCN on hydrolysis produces a white precipitate and a gas which turns filter paper moistened with copper sulphate solution deep blue.

Solution

$$3 N_2O + 4 Na + NH_3\xrightarrow[]{NH_3(\ell)}NaN_3 + 3 NaOH + 2 N_2$$
$$\underbrace { 2NH_3 +NaOCl }_{ dilute \: solutions} \xrightarrow[]{glue} N_2H_4 + NaCl + H_2O$$
$$(NH_4)_2Cr_2O_7\xrightarrow[]{\Delta }N_2 + 4 H_2O + Cr_2O_3\downarrow (green)$$
$$CaCN_2 + 3 H_2O\longrightarrow CaCO_3\downarrow +2 NH_3$$
$$Cu^{2+} + 4 NH_3\longrightarrow [Cu(NH_3)_4]^{2+} (deep \: blue)$$
Question 16
1 / -0

In the electronic structure of H$$_2$$SO$$_4$$, the total number of shared electrons is:

A
$$20$$

B
$$16$$

C
$$12$$

D
$$8$$

Solution

In the electronic structure of sulphuric acid, total $$8$$ bonds are present.
Each bond is formed by sharing of two electrons. Thus, total $$16$$ electrons are shared.
Question 17
1 / -0

Directions For Questions

A physicist wishes to eject electrons by shining light on the on a metal surface. The light source emits light of wavelength of 450 nm. The table list the only available metals and their work functions.
Metal $$W_0(eV)$$
Barium 2.5
Lithium 2.3
Tantalum 4.2
Tungston 4.5

...view full instructions

Which metal(s) can be used to produce electrons b the photoelectric effect from given source of light?

A
Barium only

B
Barium or lithium

C
Lithium, tantalum or tungsten

D
Tungsten or tantalum

Solution

$$\Delta E=\frac {12400}{5400A^o}$$
$$\Delta 2.75 eV$$
For photoelectric effect, $$\Delta E > W_0$$ (work function).
Question 18
1 / -0

Which of the following compound has a directional bond?

A
Magnesium chloride

B
Solid neon

C
Hydrogen sulphide

D
Caesium fluoride

Solution

Ionic bonds are non-directional bond or multi directional bonds. While covalent bonds are directional bond due to sharing of electrons.

There is ionic bond present in magnesium chloride and caesium fluoride.
In hydrogen sulphide, there is a presence of covalent bond.

Thus, it consists of the directional bond.

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Metal	$$W_0(eV)$$
Barium	2.5
Lithium	2.3
Tantalum	4.2
Tungston	4.5

Chemical Bonding and Molecular Structure Test -6

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Chemical Bonding and Molecular Structure Test -6

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Question 1

sp, 0

dsp2, 1

sp3, 0

sp2, 0

Solution

Question 2

CH4→ C2H6

BF3→ BF4−

NH3→ NH4+

H2O → H3O+

Solution

Question 3

2

8

4

0

Solution

Question 4

SiO2−3

SO3

BrF3

SF2O

Solution

Question 5

negative

zero

there is no relationship between stability and bond order

positive

Solution

Question 6

two sigma and one pi bonds

one sigma and two pi bonds

three pi bonds

three sigma bonds

Solution

Question 7

+ 0.75

– 0.75

– 1

+ 1

Solution

Question 8

HCl

HI

H2O

H2S

Solution

Question 9

$$1 D = 3.3356 Cm$$

$$1D= 3.3356\times10^{-30}C m$$

$$1 Cm = 3.3356 D$$

$$1 C m = 3.3356\times10^{-30} D$$

Solution

Question 10

$$H_3O^+$$

$$BF_4^-$$

$$HF_2^-$$

$$NH_4^+$$

Solution

Question 11

$$(s + p_y)$$ produces sp hybrid orbitals which are lying in the yz plane

$$(s + p_y)$$ produces sp hybrid orbitals which are lying the xz plane

$$(s + p_x + p_z)$$ produces sp2 hybrid orbitals which are lying in the xz plane

$$(s p_y)$$ produces sp hybrid orbitals which are lying along the y-axis

Solution

Question 12

$$sp > sp^{2}< sp^{3}$$

$$sp > sp^{2} > sp^{3}$$

$$sp < sp^{2}> sp^{3}$$

$$sp < sp^{2}< sp^{3}$$

Solution

dsp², 1

sp³, 0

sp², 0

CH₄→ C₂H₆

BF₃→ BF₄⁻

NH₃→ NH₄⁺

H₂O → H₃O⁺

SiO²⁻₃

SO₃

BrF₃

SF₂O

H₂O

H₂S