As the oxidation state of central atom increases the acidic character increases, so the correct order of acidic character is
$$\mathop {{P_4}}\limits^{ + 5} {O_{10}} > \mathop {{P_4}}\limits^{ + 3} {O_6} > \mathop {A{s_4}}\limits^{ + 5} {O_{10}} > \mathop {A{s_4}}\limits^{ + 3} {O_6}$$
and hence, \[A{{s}_{4}}{{O}_{6}}\] is least acidic.
624.
Graphite is a soft solid lubricant extremely difficult to melt. The reason for this anomalous behaviour is that graphite
A
is an allotropic form of diamond
B
has molecules of variable molecular masses like
polymers
C
has carbon atoms arranged in large plates of rings of
strongly bound carbon atoms with weak interplate
bonds
D
is a non-crystalline substance
Answer :
has carbon atoms arranged in large plates of rings of
strongly bound carbon atoms with weak interplate
bonds
In graphite, carbon is $$s{p^2}$$ hybridized. Each carbon is thus linked to three other carbon atoms forming hexagonal rings. Since only three electrons of each carbon are used in making hexagonal ring, fourth electron of each carbon is free to move. This makes graphite a good conductors of heat and electricity. Further graphite has a two dimensional sheet like structure. These various sheets are held together by weak van der Waal’s force of attraction. due to these weak forces of attraction, one layer can slip over the other. This makes graphite soft and a good lubricating agent.
625.
Which of the following group - 14 elements is a radioactive element?
The order of boiling point in group 18 is $$He < Ne < Ar < Kr < Xe$$
628.
Which of the following statements are correct ?
(i) Aluminium forms $${\left[ {Al{F_6}} \right]^{3 - }}ion$$ while boron forms only $${\left[ {B{F_4}} \right]^ - }\,ion$$ due to presence of $$d$$ - orbitals in aluminium.
(ii) The first member of a group differs from the heavier members in its ability to form $$p\pi {\text{ - }}p\pi $$ multiple bonds to itself and to other second row elements. While heavier member forms $$d\pi {\text{ - }}p\pi $$ bonds.
(iii) $$d$$ - orbitals contribute more to the overall stability of molecules than $$p\pi {\text{ - }}p\pi $$ bonding of second row elements.
$$d$$ - orbitals are of higher energy than the $$p$$ - orbitals, they contribute less to the overall stability of molecules than $$p\pi {\text{ - }}p\pi $$ bonding of the second row elements.
629.
An inorganic compound $$'A'$$ shows the following reactions :
(i) It is white solid, exists as dimer and fumes in wet air.
(ii) It sublimes at $$180{\,^ \circ }C$$ and forms monomer if heated to $$400{\,^ \circ }C.$$
(iii) Its aqueous solution turns blue litmus to red and gives a white precipitate with $$AgN{O_3}$$ solution, which is soluble in $$N{H_4}OH.$$
(iv) Addition of $$N{H_4}OH$$ and $$NaOH$$ separately to the solution of $$'A'$$ gives a gelatinous precipitate which in however soluble in excess of $$NaOH.$$
The compound $$'A'$$ is
Compound $$'A'$$ is $$A{l_2}C{l_6}.$$
(i) $$A{l_2}C{l_6}$$ is a white solid, exists as dimer and fumes in wet air.
(ii) It sublimes at $$180{\,^ \circ }C$$ and converts into $$AlC{l_3}$$ at $$400{\,^ \circ }C.$$
(iii) Its aqueous solution is acidic, so it turns blue litmus to red.
$$A{l_2}C{l_6} + 6{H_2}O \to $$ $$\mathop {2Al{{\left( {OH} \right)}_3}}\limits_{{\text{weak base}}} + \mathop {6HCl}\limits_{{\text{strong acid}}} $$
It gives white $$ppt.$$ with $$AgN{O_3}$$ which is soluble in $$N{H_4}OH.$$
$$\left( {{\text{iv}}} \right)AlC{l_3} + 3NaOH \to $$ $$\mathop {Al{{\left( {OH} \right)}_3}}\limits_{{\text{white ppt}}{\text{.}}} + 3NaCl$$
$$Al{\left( {OH} \right)_3} + \mathop {NaOH}\limits_{{\text{excess}}} \to $$ $$\mathop {NaAl{O_2}}\limits_{{\text{soluble}}} + 2{H_2}O$$
But in case of $$N{H_4}OH,$$ the $$ppt.$$ of $$Al{\left( {OH} \right)_3}$$ does not dissolve in excess of $$N{H_4}OH.$$
630.
The species, having bond angles of $${120^ \circ }$$ is
The species having bond angles of $${120^ \circ }$$ is $$BC{l_3}.$$
It is $$s{p^2}$$-hybridised and central atom does not have any lone pair of electrons.