JEE 2012 :: General Questions :: Chemistry

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$NiCl_{2}[P(C_{2}H_{5})_{2}(C_{6}H_{5})]_{2}$ exhibits temperature dependent magnetic behaviour (paramagnetic/diamagnetic) the coordination geometries of $Ni^{2+}$ in the paramagnetic and diamagnetic states are respectively 

Answer:

Explanation:

In the given complex, $NiCl_{2}[P(C_{2}H_{5})_{2}(C_{6}H_{5})]_{2}$  nickel is in + 2 oxidation state and the ground state electronic configuration of  $Ni^{2+}$ ion in the free gaseous state is

For the given four-coordinated complex to be paramagnetic, it must Possess unpaired electrons in the valence shell. To satisfy this condition, four lone pairs from the four ligands occupy the four  $sp^{3}$ hybrid orbital as,

Therefore, the Geometry of the paramagnetic complex must be tetrahedral. On the other hand, for the complex to be diamagnetic, there should not be any unpaired electrons in the valence shell. This condition can be fulfilled by pairing electrons of 3d orbitals against Hund's rule as

The above electronic arrangements give dsp² hybridizations and therefore square-planar geometry to the complex

29.2% (w/W) HCl, stock solution has a density of 1.25 g mL^-1. The molecular weight of HCl is 36.5 g mol^-1. The volume (mL) of stock solution required to prepare a 200 mL solution 0.4 M HCl is

Answer:

Explanation:

Mass of HCl in 1.0 mL stock solution

   $=1.25 \times \frac{29.2}{100}$=0.365 g

Mass of HCl required for 200 mL  0.4 M HCl

= $\frac{200}{100} \times 0.4 \times 36.5=0.08 \times 36.5g$

$\therefore$   0.365g of HCl is present on 1.0mL stock solution

$\therefore$   $0.08 \times 36.5$ g HCl will be present in $\frac{0.08 \times 36.5}{0.365}=8.0mL$

The Periodic Table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group, element X belongs in the Periodic Table?

$_{29}^{63}Cu+_{1}^{1}H-6_{0}^{1}n+_{2}^{4}\alpha+2_{1}^{1}H+X$

Answer:

Explanation:

Balancing the given nuclear reaction in terms of atomic number (charge) and mass number gives

$_{29}Cu^{63}+_{1}H^{1}\rightarrow6_{0}n^{1}+_{2}He^{4}(a)+2_{1}H^{1}+_{2}He^{4}+_{26}X^{52}$

The atomic number 26 correspond to transition metal "Fe" which belongs to 8^th group of modern Periodic Table

The substituents $R_{1}$ and $R_{2}$ for nine peptides are listed in the table given below. How many of these peptides are positively charged at pH=7.0?

Answer:

Explanation:

The amino acid remains completely in Zwitterionic form at its isoelectric point. Amino acids with the additional acidic group have their isoelectric pH less than 7.0 and increasing pH above the isoelectric  point makes them anionic. On the other hand, amino acids with the additional basic groups have their isoelectric pH greater than 7.0 and decreasing pH below the isoelectric point (by adding acid solution) makes them cationic. The given peptide with following R₁ and R₂ are basic, will remain Protonated (cationic) at PH = 7.0.

When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its pyranose form is

Answer:

Explanation:

The D-from of given sugar is

Configurations at the three chiral carbons (starred) can be changed maintaining D-configuration. Hence, the total number of stereoisomers of D-pyranose=$2^{3}=8$

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