P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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591.
Which of the following is a tetrabasic acid?
A
Hypophosphorous acid
B
Metaphosphoric acid
C
Pyrophosphoric acid
D
Orthophosphoric acid
Answer :
Pyrophosphoric acid
No explanation is given for this question. Let's discuss the answer together.
592.
How many $$P - O - P$$ bonds appear in cyclotrimetaphosphoric acid?
A
Four
B
Three
C
Two
D
One
Answer :
Three
Cyclotrimetaphosphoric acid is $${\left( {HP{O_3}} \right)_3}.$$
593.
Which of the following shows nitrogen in its increasing order of oxidation number?
$${N_2}O = + 1,NO = + 2,N{O_2} = + 4,$$ $$NO_3^ - = + 5,NH_4^ + = - 3$$
Increasing order of oxidation state will be
$$NH_4^ + < {N_2}O < NO < N{O_2} < NO_3^ - $$
594.
Thenumber of $$S - S\,$$ bonds in sulphur trioxide trimer $$\left( {{S_3}{O_9}} \right)$$ is
A
three
B
two
C
one
D
zero
Answer :
zero
In sulphur trioxide trimer $${{S_3}{O_9}}$$ ( also called $$\gamma - $$ sulphur trioxide ) two sulphur atoms are linked to each other via $$O\,$$ atoms, hence there is no $$S - S$$ bond.
595.
Which of the following on thermal decomposition gives oxygen gas ?
A
$$A{g_2}O$$
B
$$P{b_3}{O_4}$$
C
$$Pb{O_2}$$
D
$${\text{All of these}}$$
Answer :
$${\text{All of these}}$$
$$\eqalign{
& 2A{g_2}O\left( s \right) \to 4Ag\left( s \right) + {O_2}\left( g \right) \cr
& 2P{b_3}{O_4}\left( s \right) \to 6PbO\left( s \right) + {O_2}\left( g \right) \cr
& 2Pb{O_2}\left( s \right) \to 2PbO\left( s \right) + {O_2}\left( g \right) \cr} $$
597.
The tendency of $$B{F_3},BC{l_3}$$ and $$BB{r_3}$$ behave as Lewis acid decreases in the sequence
A
$$BC{l_3} > B{F_3} > BB{r_3}$$
B
$$BB{r_3} > BC{l_3} > B{F_3}$$
C
$$BB{r_3} > B{F_3} > BC{l_3}$$
D
$$B{F_3} > BC{l_3} > BB{r_3}$$
Answer :
$$BB{r_3} > BC{l_3} > B{F_3}$$
As the size of halogen atom increases, the acidic strength of boron halides increases. Thus, $$B{F_3}$$ is the weakest Lewis acid. This is because of the $$p\pi - p\pi $$ back bonding between the fully filled unutilised $$2p$$ -orbitals of $$F$$ and vacant $$2p$$ -orbitals of boron which makes $$B{F_3}$$ less electron deficient. Such back donation is not possible in case of $$BC{l_3}$$ or $$BB{r_3}$$ due to larger energy difference between their orbitals. Thus, these are more electron deficient. Since on moving down the group the energy difference increases, the Lewis acid character also increases. Thus, the tendency to behave as Lewis acid follows the order
$$BB{r_3} > BC{l_3} > B{F_3}$$
598.
Carbon shows a maximum covalency of four whereas other members can expand their covalence. It is because of
A
absence of $$d$$ - orbitals in carbon
B
ability of carbon to form $$p\pi - p\pi $$ multiple bonds
C
small size of carbon
D
catenation of carbon
Answer :
absence of $$d$$ - orbitals in carbon
No explanation is given for this question. Let's discuss the answer together.
599.
Covalency of oxygen cannot exceed 2 unlike sulphur which can show + 4 or + 6 because
A
oxygen atom does not have $$d$$ - orbitals
B
oxygen atom has two unpaired electrons in its valence shell
C
oxygen can form a double bond with another oxygen atom
D
electrons of oxygen atom cannot be promoted to $$d$$ - orbitals due to its small size
Answer :
oxygen atom does not have $$d$$ - orbitals
Valence shell electronic configuration of oxygen is $$2{s^2}2{p^4}.$$
600.
Which of the following is most acidic?
A
$${N_2}{O_5}$$
B
$${P_2}{O_5}$$
C
$$A{s_2}{O_5}$$
D
$$S{b_2}{O_5}$$
Answer :
$${N_2}{O_5}$$
Acidic nature of oxides decreases down a group.
So, $${N_2}{O_5}$$ is most acidic.
Another reason of acidic strength of $${N_2}{O_5}$$ is that the electronegativity of $$N$$ is maximum in the given $${V^{th}}$$ group elements. As we know that on increasing the electronegative character, acidic nature increases.