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The number of geometric isomers that can exit for square planar   $ [Pt(Cl)(py) (NH_{3})(NH_{2}OH)]^{+}$   is [py= pyridine]

Answer:

Explanation:

$ [Pt(Cl)(py) (NH_{3})(NH_{2}OH)]^{+}$   is square planar complex.

 The structures are formed by fixing a group and then arranging all groups

 Hence, this complex shows geometrical isomers.

Which one has the highest boiling point ?

Answer:

Explanation:

As we move down the group of noble gases, molecular mass increases by which dipole produced for a moment and hence London forces increases from He to Xe . Therefore more amount of energy is required to break these forces, thus the boiling point also increases from He to Xe.

Match the catalysts to the correct processes.

       Catalyst                                         Process

(A)  $TiCl_{3}$                                (i)  Wacker process

(B)   $PdCl_{2}$                              (ii)   Ziegler-Natta Polymerisation

(C)   $CuCl_{2}$                              (iii)   Contact process

(D)  $V_{2}O_{5}$                          (iV)  Deacon's  process

Answer:

Explanation:

(a)   $TiCl_{3}$   is used as Ziegler -Natta catalyst for the polymerisation of ethene.

(b)  $PdCl_{2}$   is used in the Wacker process, in which alkene changed into aldehyde via catalytic cyclic process initiated by  $PdCl_{2}$.

(c)   $CuCl_{2}$ is used in Deacon's process (for Cl₂)

(d)  $V_{2}O_{5}$    is used in contact process of manufacturing sulphuric acid.

 Time Saving Technique.

                    This type of questions can also be solved through elimination technique. There is no need to know all four matches to select the correct response. Even if you know (b) matches then also you can solve the problem. e.g,  suppose you know the usage of  $V_{2}O_{5}$ in contact process (i,e,  D matches with  (iii)  and  $TiCl_{3}$ is connected to Ziegler-Natta catalyst  (i.e,  A matches with ii).  These two combinations are present only in option number (b). Likewise, for this question particularly if you know that $V_{2}O_{5}$ is used in contact process then this combination is present in option (b) only out of all the four option given. In this way, you can eliminate wrong options to get the correct response.

Which among the following is the most reactive?

Answer:

Explanation:

$Cl_{2}$,  $Br_{2}$ and  $I_{2}$ are homonuclear diatomic molecule in which electronegativity of the combining atoms is same, so they are more stable and less reactive, whereas, I and Cl have different electronegativites and bond between them are polarised and hence, reactive. Therefore, interhalogen compounds are more reactive.

 Time Saving Technique.

In this type of question of halogen, only go through the polarity of the molecule. As we know, diatomic molecule does not have polarity but molecules with dissimilar sizes have polarity resulting in more reactivity.

Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?

Answer:

Explanation:

 As we move down the group, size of metal increases. Be has lower  size while $SO_{4}^{2-}$ has  bigger size, that's why  

$BeSO_{4}$ breaks easily and lattice energy becomes smaller but due to lower size of Be, water molecules are gathered around and hence hydration energy increases.

  On the other hand, rest metals i.e, Ca, Ba, Sr have bigger size and that's why lattice energy is greater than hydration energy.

Time-saving Technique.

   In the question of finding hydration energy only check the size of atom. Smaller sized atom has more hydration energy.

  Thus, in this question, Be is placed uppermost in the group has lesser size and not  comparable with the size of sulphates. Hence ,   $BeSO_{4}   is right response.

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• Main 2015 - Chemistry 2015
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