The p-Block Elements Test

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The p-Block Elements Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

Inert gases are mixed in iodine vapours. Then there are ________ between them.

A
H-bonding

B
Vander Waal's forces

C
Electrostatic forces

D
Metallic bonds

Solution

Noble gases are least reactive. Their inertness to chemical reactivity is beacuse of the completely filled $$ns^2np^6$$
electronic configuration in their valence shell except helium.
Hence, their only exists the short range force i.e. Vander waal's force between them.
Question 2
1 / -0

The strongest oxidising agent is :

A
$$F_2$$

B
$$Cl_2$$

C
$$I_2$$

D
$$Br_2$$

Solution

An element which can readily accept an electron is a good oxidising agent. Halogens are good oxidising agents because they have strong tendency to accept an electron due to $$(i)$$ small atomic size, $$(ii)$$ high electronegativity and $$(iii)$$ high electron affinity. Fluorine, being smallest, is best oxidising agent.
Question 3
1 / -0

The correct order of solubility in water for He, Ne, Ar, Kr, Xe is :

A
$$He > Ne > Ar > Kr > Xe$$

B
$$Xe > Kr > Ar > Ne > He$$

C
$$Ne > Ar > Kr > He > Xe$$

D
$$Ar > Ne > He > Kr > Xe$$

Solution

As the molecular weight of noble gas atoms increases down the group its polarity increases due to which van-der-waal's force between them increases. Due to increased polarity of heavier inert gas, its solubility in water also increases. So, most soluble gas will be $$Xe$$ and least soluble will be $$He$$. So correct order is $$Xe > Kr > Ar > Ne He$$.
Question 4
1 / -0

The oxidation number of chlorine in KClO is:

A
+ 1

B
+ 3

C
+ 5

D
+ 7

Solution

$$\overset{+1}{K}Cl\overset{-2}{O}$$
oxidation state of alrali metals = +1
oxygen = -2
Let oxidation state of chlorine is x.
$$\therefore$$ x+1+(-2) = 0
$$\therefore$$ x = +1
Thus, oxidation state of chlorine in potassium chlorate is +1.
Question 5
1 / -0

Assertion (A): Oxygen exhibits anomalous behavior.
Reason (R): Absence of d-orbitals

A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion

B
Both Assertion and Reason are true and Reason is not the correct explanation of Assertion

C
Assertion is true but Reason is false

D
Assertion is false but Reason is true

Solution

Oxygen exhibits anomalous behaviour due to its small size, high electronegativity and absence of d-orbitals.
Question 6
1 / -0

Anomalous behaviour of oxygen, compared to other VIA group elements is due to:

A
its high electronegativity

B
its small atomic size

C
non availability of d-orbitals

D
all

Solution

Oxygen is the first element of VI A group and shows Anaomalous behaviour than other group elements because
1:High Electronegativity
2:small atomic size
3:Non-avalability of d-orbits results in different compounds than their other group elements.
Question 7
1 / -0

Fluorine shows anomalous behavior in VIIA group due to:

A
its small size

B
high electronegativity

C
absence of d-orbitals

D
All

Solution

Second period elements show anomalous behavior due to small size, high electronegativity and absence of d-orbitals and fluorine is a second period element
Question 8
1 / -0

According to Coulson the high reactivity of fluorine than other halogens is because of its:

A
high electronegativity

B
lesser internuclear repulsions

C
greater repulsions between non bonding electrons

D
All the above

Solution

According to Coulson, since fluorine atoms are small it's intermolecular repulsion is appreciable. The large electron-electron repulsion between the lone pairs of electrons on the fluorine atom weakens the bond.
Question 9
1 / -0

Which of the following halogens does not form oxyacids at room temperature?

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

B
$$ \mathrm{Cl}_{2}$$

C
$$\mathrm{B}\mathrm{r}_{2}$$

D
$$I_{2}$$

Solution

Fluorine does not hove d-sub shell so it can only have one oxidation state. Therefore, it cann't form oxy acids. Fluorine is the most electronegative element and always show $$-1$$ oxiadation state. In oxyacids. reast of the halogens have positive oxidation state.
Question 10
1 / -0

The number of p-electrons in bromine atom is:

A
17

B
7

C
15

D
13

Solution

$$Br - 35$$
Its electronic configuration is given as:
$$1s^22s^22p^63s^23p^64s^23d^{10}4p^5$$
So, total p-electrons =6+6+5
= 17

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 18

Result

The p-Block Elements Test - 18

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SHARING IS CARING

Question 1

H-bonding

Vander Waal's forces

Electrostatic forces

Metallic bonds

Solution

Question 2

$$F_2$$

$$Cl_2$$

$$I_2$$

$$Br_2$$

Solution

Question 3

$$He > Ne > Ar > Kr > Xe$$

$$Xe > Kr > Ar > Ne > He$$

$$Ne > Ar > Kr > He > Xe$$

$$Ar > Ne > He > Kr > Xe$$

Solution

Question 4

+ 1

+ 3

+ 5

+ 7

Solution

Question 5

Both Assertion and Reason are true and Reason is the correct explanation of Assertion

Both Assertion and Reason are true and Reason is not the correct explanation of Assertion

Assertion is true but Reason is false

Assertion is false but Reason is true

Solution

Question 6

its high electronegativity

its small atomic size

non availability of d-orbitals

all

Solution

Question 7

its small size

high electronegativity

absence of d-orbitals

All

Solution

Question 8

high electronegativity

lesser internuclear repulsions

greater repulsions between non bonding electrons

All the above

Solution

Question 9

$$\mathrm{F}_{2}$$

$$ \mathrm{Cl}_{2}$$

$$\mathrm{B}\mathrm{r}_{2}$$

$$I_{2}$$

Solution

Question 10

17

7

15

13

Solution

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