Tag: chemistry

Questions Related to chemistry

Which of the following exists as associated molecules due to hydrogen bonding?

  1. $NH _{3}$

  2. $H _{2}O$

  3. $HF$

  4. $HCl$


Correct Option: A,B,C
Explanation:

Due to good extent of hydrogen bonding (hydrogen bonding is more favoured if the hydrogen is attached to an electronegative atom) in A, B and C options, they exist in chains or in groups attached through hydrogen bonding.

Match list I with list II. Choose the correct matching codes from the choices given. 
     List I                                                 List II
   (Hydride)                                    (Type of hydride)
A. $BeH _{2}$                        1.  Complex
B.. $AsH _{3}$                       2. Lewis acid
C. $B _{2}H _{6}$                   3. Interstitial
D. $LaH _{3}$                        4. Covalent
E. $LiAlH _{4}$                      5. Intermediate

  1. A-6, B-2, C-4, D-5, E-1

  2. A-6, B-2, C-4, D-3, E-1

  3. A-6, B-4, C-2, D-3, E-15

  4. A-5, B-4, C-2, D-3, E-1


Correct Option: D
Explanation:

$A.$ $BeH _2\rightarrow$ Intermediate

$B.$ $AsH _3\rightarrow$ Covalent
$C.$ $B _2H _6\rightarrow$ Lewis acid
$D.$ $LaH _3\rightarrow$ Interstitial
$E.$ $LiAlH _4\rightarrow$ Complex

Match the column I with column II and mark the appropriate choice.

column I column II
A NaH i Interstitial hydride 
B $CH _{4}$ ii Molecular hydride
C $VH _{0.56}$ iii Ionic hydride 
D $B _{2}H _{6}$ iv  Electron-deficient hydride 
  1. $(A)\rightarrow (iii), (B)\rightarrow (iv), (C)\rightarrow (ii) (D)\rightarrow (i)$

  2. $(A)\rightarrow (ii), (B)\rightarrow (iv), (C)\rightarrow (iii) (D)\rightarrow (i)$

  3. $(A)\rightarrow (i), (B)\rightarrow (ii), (C)\rightarrow (iv) (D)\rightarrow (iii)$

  4. $(A)\rightarrow (iii), (B)\rightarrow (ii), (C)\rightarrow (i) (D)\rightarrow (iv)$


Correct Option: D
Explanation:
NaH is an ionic compound and is made of sodium cations $\left ( Na^{+} \right )$ and hydride anions $\left ( H^{-} \right )$. It has the octahedral crystal structure with each sodium ion surrounded by six hydride ions.

$CH _{4}$ forms molecular or covalent hydrides. These are mainly formed by p - block elements and some s - block elements.

The third compound has an oxidation number of hydrogen which is zero. So it belongs to Interstitial hydride.

In $B _{2}H _{6}$ the Boron atom is surrounded by 6 electrons, so it is electron deficient due to its incomplete octet.

The correct option is D.

Correct order of ionic character is:

  1. $CaH _2 < BaH _2 < CH _4$

  2. $CH _4 < CaH _2 < BaH _2$

  3. $CH _4 < BaH _2 < CaH _2$

  4. $BaH _2< CaH _2 < CH _4$


Correct Option: C

Ionic hydride among following is 

  1. $BeH _2$

  2. $NH _2$

  3. $CaH _2$

  4. $CuH _2$


Correct Option: C

Hydrogen differ with alkali metals in the following aspect ?

  1. Formation of halides

  2. Formation of oxides

  3. Electronic configuration

  4. Atomicity


Correct Option: A

Among $CaH _2, NH _3, NaH$ and $B _2H _6$, which are covalent hydrides?

  1. $NH _3$ and $B _2H _6$

  2. $NaH$ and $CaH _2$

  3. $NaH$ and $ NH _3$

  4. $CaH _2$ and $B _2H _6$


Correct Option: A
Explanation:

$NH _3$ and $B _2H _6$ are the covalent hydrides and $CaH _2$ and $NaH$ are the ionic hydride.

$N$ and $B$ form covalent hydride as they have high electronegativity and small size.

Hydrogen is:

  1. electropositive

  2. electronegative

  3. both electropositive as well as electronegative

  4. neither electropositive nor electronegative


Correct Option: C
Explanation:
Hydrogen is both electropositive as well as electronegative, as it forms hydrogen ion $H^+$ as well as hydride ion $H^-$.

Hydrogen readily combines with these elements?

  1. Na

  2. K

  3. Ca

  4. Zn


Correct Option: A,B,C
Explanation:

${ H } _{ 2 }$ easily reacts with $Na, Ca, K$ as they are highly reactive metals and are placed at the top of the reactivity series.

${ 2Na }+{ H } _{ 2 }\rightarrow 2NaH$
$2K+{ H } _{ 2 }\rightarrow 2KH$
$Ca+{ H } _{ 2 }\rightarrow CaH _2$.

Which of the following is not correctly matched?

  1. $PCl _{5}$ - $sp^{3}d$ hybridisation.

  2. $PCl _{3}$ - $sp^3$ hybridisation.

  3. $PCl _{5}$ (solid) - $[PtCl _{4}]^{+}$+ $[PtCl _{6}]^{-}$.

  4. $PCl _{5}$ - brownish powder.


Correct Option: D
Explanation:


 Colour of $PCl _5$ is not brownish but a greenish yellow crystalline solid with an irritating odour.

Hence option D is correct.