274179
For the crystal field splitting in octahedral complexes,
1 the energy of the $e_{g}$ orbitals will decrease by $(3 / 5) \Delta$ 。 and that of the $t_{2 g}$ will increase by $(2 / 5) \Delta$.
2 the energy of the $e_{g}$ orbitals will increase by $(3 / 5) \Delta$ 。 and that of the $t_{2 g}$ will decrease by $(2 / 5) \Delta$ 。
3 the energy of the $e_{g}$ orbitals will increase by $(3 / 5) \Delta_{\circ}$, and that of the $t_{2 g}$ will increase by $(2 / 5) \Delta$ 。
4 the energy of the $e_{g}$ orbitals will decrease by $(3 / 5) \Delta_{\circ}$, and that of the $t_{2 g}$ will decrease by $(2 / 5) \Delta$ 。
Explanation:
(B) : In octahedral complexes, the d-orbital splitted in two degenerate set i.e. $t_{2 g}$ and $e_{g} . t_{2 g}$ has the lower energy than the $\mathrm{e}_{\mathrm{g}}$ orbital.
Kerala-CEE-29.08.2021
COORDINATION COMPOUNDS
274180
Ethylene diaminetetraacetate (EDTA) ion is
1 tridentate ligand with three "N" donor atoms
2 hexadentate ligand with four "O" and two "N" donor atoms
3 unidentate ligand
4 bidentate ligand with two "N" donor atoms.
Explanation:
(B) : Ethylene diaminetetra acetate (EDTA) ion is 6 donor atoms so it is called hexadentate ligand with four "O" and two "N" donor atoms.
NEET 2021
COORDINATION COMPOUNDS
274185
Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements?
1 P $<$ N $<$ F $<$ Cl
2 N $<$ P $<$ F $<$ Cl
3 Cl $<$ F $<$ P $<$ N
4 $\mathrm{L}<\mathrm{Cl}<\mathrm{N}<$ P
Explanation:
(B) : As a general rule, electron gain enthalpy becomes more negative with increase in the atomic number across a period Nitrogen and Fluorine are the element of second period and phosphorus and chlorine are prom the third period. The size of nitrogen are less to the phosphorous atom As we know halogens have more negative electron gain enthalpy. So the order will be - $\mathrm{N}<\mathrm{P}<\mathrm{F}<\mathrm{Cl}$.
AP EAPCET 19-08-2021 Shift-I
COORDINATION COMPOUNDS
274188
For $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ and $\mathrm{CuCl}_{2}$ in aqueous medium, which of the following statement is correct?
1 $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$
2 Stability of $\mathrm{Cu}_{2} \mathrm{Cl}_{2} \mathrm{i}$ equal to stability of $\mathrm{CuCl}_{2}$
3 Both are unstable
4 $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ is more stable than $\mathrm{CuCl}_{2}$
Explanation:
(A) : $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ in aqueous medium because stability in aqueous medium depends and the hydration enthalpy of ion. Hydration enthalpy depends upon the size of ions. Hydration enthalpy $\propto \frac{1}{\text { Size of ions }}$ In $\mathrm{CuCl}_{2}$ copper is present in +2 oxidation and in $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ oxidation state is +1 size of $\mathrm{Cu}^{2+}$ is smaller than $\mathrm{Cu}^{+}$therefore hydration energy of $\mathrm{Cu}^{2+}$ is higher than $\mathrm{Cu}^{+}$. Hence, $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$.
274179
For the crystal field splitting in octahedral complexes,
1 the energy of the $e_{g}$ orbitals will decrease by $(3 / 5) \Delta$ 。 and that of the $t_{2 g}$ will increase by $(2 / 5) \Delta$.
2 the energy of the $e_{g}$ orbitals will increase by $(3 / 5) \Delta$ 。 and that of the $t_{2 g}$ will decrease by $(2 / 5) \Delta$ 。
3 the energy of the $e_{g}$ orbitals will increase by $(3 / 5) \Delta_{\circ}$, and that of the $t_{2 g}$ will increase by $(2 / 5) \Delta$ 。
4 the energy of the $e_{g}$ orbitals will decrease by $(3 / 5) \Delta_{\circ}$, and that of the $t_{2 g}$ will decrease by $(2 / 5) \Delta$ 。
Explanation:
(B) : In octahedral complexes, the d-orbital splitted in two degenerate set i.e. $t_{2 g}$ and $e_{g} . t_{2 g}$ has the lower energy than the $\mathrm{e}_{\mathrm{g}}$ orbital.
Kerala-CEE-29.08.2021
COORDINATION COMPOUNDS
274180
Ethylene diaminetetraacetate (EDTA) ion is
1 tridentate ligand with three "N" donor atoms
2 hexadentate ligand with four "O" and two "N" donor atoms
3 unidentate ligand
4 bidentate ligand with two "N" donor atoms.
Explanation:
(B) : Ethylene diaminetetra acetate (EDTA) ion is 6 donor atoms so it is called hexadentate ligand with four "O" and two "N" donor atoms.
NEET 2021
COORDINATION COMPOUNDS
274185
Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements?
1 P $<$ N $<$ F $<$ Cl
2 N $<$ P $<$ F $<$ Cl
3 Cl $<$ F $<$ P $<$ N
4 $\mathrm{L}<\mathrm{Cl}<\mathrm{N}<$ P
Explanation:
(B) : As a general rule, electron gain enthalpy becomes more negative with increase in the atomic number across a period Nitrogen and Fluorine are the element of second period and phosphorus and chlorine are prom the third period. The size of nitrogen are less to the phosphorous atom As we know halogens have more negative electron gain enthalpy. So the order will be - $\mathrm{N}<\mathrm{P}<\mathrm{F}<\mathrm{Cl}$.
AP EAPCET 19-08-2021 Shift-I
COORDINATION COMPOUNDS
274188
For $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ and $\mathrm{CuCl}_{2}$ in aqueous medium, which of the following statement is correct?
1 $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$
2 Stability of $\mathrm{Cu}_{2} \mathrm{Cl}_{2} \mathrm{i}$ equal to stability of $\mathrm{CuCl}_{2}$
3 Both are unstable
4 $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ is more stable than $\mathrm{CuCl}_{2}$
Explanation:
(A) : $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ in aqueous medium because stability in aqueous medium depends and the hydration enthalpy of ion. Hydration enthalpy depends upon the size of ions. Hydration enthalpy $\propto \frac{1}{\text { Size of ions }}$ In $\mathrm{CuCl}_{2}$ copper is present in +2 oxidation and in $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ oxidation state is +1 size of $\mathrm{Cu}^{2+}$ is smaller than $\mathrm{Cu}^{+}$therefore hydration energy of $\mathrm{Cu}^{2+}$ is higher than $\mathrm{Cu}^{+}$. Hence, $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
COORDINATION COMPOUNDS
274179
For the crystal field splitting in octahedral complexes,
1 the energy of the $e_{g}$ orbitals will decrease by $(3 / 5) \Delta$ 。 and that of the $t_{2 g}$ will increase by $(2 / 5) \Delta$.
2 the energy of the $e_{g}$ orbitals will increase by $(3 / 5) \Delta$ 。 and that of the $t_{2 g}$ will decrease by $(2 / 5) \Delta$ 。
3 the energy of the $e_{g}$ orbitals will increase by $(3 / 5) \Delta_{\circ}$, and that of the $t_{2 g}$ will increase by $(2 / 5) \Delta$ 。
4 the energy of the $e_{g}$ orbitals will decrease by $(3 / 5) \Delta_{\circ}$, and that of the $t_{2 g}$ will decrease by $(2 / 5) \Delta$ 。
Explanation:
(B) : In octahedral complexes, the d-orbital splitted in two degenerate set i.e. $t_{2 g}$ and $e_{g} . t_{2 g}$ has the lower energy than the $\mathrm{e}_{\mathrm{g}}$ orbital.
Kerala-CEE-29.08.2021
COORDINATION COMPOUNDS
274180
Ethylene diaminetetraacetate (EDTA) ion is
1 tridentate ligand with three "N" donor atoms
2 hexadentate ligand with four "O" and two "N" donor atoms
3 unidentate ligand
4 bidentate ligand with two "N" donor atoms.
Explanation:
(B) : Ethylene diaminetetra acetate (EDTA) ion is 6 donor atoms so it is called hexadentate ligand with four "O" and two "N" donor atoms.
NEET 2021
COORDINATION COMPOUNDS
274185
Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements?
1 P $<$ N $<$ F $<$ Cl
2 N $<$ P $<$ F $<$ Cl
3 Cl $<$ F $<$ P $<$ N
4 $\mathrm{L}<\mathrm{Cl}<\mathrm{N}<$ P
Explanation:
(B) : As a general rule, electron gain enthalpy becomes more negative with increase in the atomic number across a period Nitrogen and Fluorine are the element of second period and phosphorus and chlorine are prom the third period. The size of nitrogen are less to the phosphorous atom As we know halogens have more negative electron gain enthalpy. So the order will be - $\mathrm{N}<\mathrm{P}<\mathrm{F}<\mathrm{Cl}$.
AP EAPCET 19-08-2021 Shift-I
COORDINATION COMPOUNDS
274188
For $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ and $\mathrm{CuCl}_{2}$ in aqueous medium, which of the following statement is correct?
1 $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$
2 Stability of $\mathrm{Cu}_{2} \mathrm{Cl}_{2} \mathrm{i}$ equal to stability of $\mathrm{CuCl}_{2}$
3 Both are unstable
4 $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ is more stable than $\mathrm{CuCl}_{2}$
Explanation:
(A) : $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ in aqueous medium because stability in aqueous medium depends and the hydration enthalpy of ion. Hydration enthalpy depends upon the size of ions. Hydration enthalpy $\propto \frac{1}{\text { Size of ions }}$ In $\mathrm{CuCl}_{2}$ copper is present in +2 oxidation and in $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ oxidation state is +1 size of $\mathrm{Cu}^{2+}$ is smaller than $\mathrm{Cu}^{+}$therefore hydration energy of $\mathrm{Cu}^{2+}$ is higher than $\mathrm{Cu}^{+}$. Hence, $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$.
274179
For the crystal field splitting in octahedral complexes,
1 the energy of the $e_{g}$ orbitals will decrease by $(3 / 5) \Delta$ 。 and that of the $t_{2 g}$ will increase by $(2 / 5) \Delta$.
2 the energy of the $e_{g}$ orbitals will increase by $(3 / 5) \Delta$ 。 and that of the $t_{2 g}$ will decrease by $(2 / 5) \Delta$ 。
3 the energy of the $e_{g}$ orbitals will increase by $(3 / 5) \Delta_{\circ}$, and that of the $t_{2 g}$ will increase by $(2 / 5) \Delta$ 。
4 the energy of the $e_{g}$ orbitals will decrease by $(3 / 5) \Delta_{\circ}$, and that of the $t_{2 g}$ will decrease by $(2 / 5) \Delta$ 。
Explanation:
(B) : In octahedral complexes, the d-orbital splitted in two degenerate set i.e. $t_{2 g}$ and $e_{g} . t_{2 g}$ has the lower energy than the $\mathrm{e}_{\mathrm{g}}$ orbital.
Kerala-CEE-29.08.2021
COORDINATION COMPOUNDS
274180
Ethylene diaminetetraacetate (EDTA) ion is
1 tridentate ligand with three "N" donor atoms
2 hexadentate ligand with four "O" and two "N" donor atoms
3 unidentate ligand
4 bidentate ligand with two "N" donor atoms.
Explanation:
(B) : Ethylene diaminetetra acetate (EDTA) ion is 6 donor atoms so it is called hexadentate ligand with four "O" and two "N" donor atoms.
NEET 2021
COORDINATION COMPOUNDS
274185
Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements?
1 P $<$ N $<$ F $<$ Cl
2 N $<$ P $<$ F $<$ Cl
3 Cl $<$ F $<$ P $<$ N
4 $\mathrm{L}<\mathrm{Cl}<\mathrm{N}<$ P
Explanation:
(B) : As a general rule, electron gain enthalpy becomes more negative with increase in the atomic number across a period Nitrogen and Fluorine are the element of second period and phosphorus and chlorine are prom the third period. The size of nitrogen are less to the phosphorous atom As we know halogens have more negative electron gain enthalpy. So the order will be - $\mathrm{N}<\mathrm{P}<\mathrm{F}<\mathrm{Cl}$.
AP EAPCET 19-08-2021 Shift-I
COORDINATION COMPOUNDS
274188
For $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ and $\mathrm{CuCl}_{2}$ in aqueous medium, which of the following statement is correct?
1 $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$
2 Stability of $\mathrm{Cu}_{2} \mathrm{Cl}_{2} \mathrm{i}$ equal to stability of $\mathrm{CuCl}_{2}$
3 Both are unstable
4 $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ is more stable than $\mathrm{CuCl}_{2}$
Explanation:
(A) : $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ in aqueous medium because stability in aqueous medium depends and the hydration enthalpy of ion. Hydration enthalpy depends upon the size of ions. Hydration enthalpy $\propto \frac{1}{\text { Size of ions }}$ In $\mathrm{CuCl}_{2}$ copper is present in +2 oxidation and in $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$ oxidation state is +1 size of $\mathrm{Cu}^{2+}$ is smaller than $\mathrm{Cu}^{+}$therefore hydration energy of $\mathrm{Cu}^{2+}$ is higher than $\mathrm{Cu}^{+}$. Hence, $\mathrm{CuCl}_{2}$ is more stable than $\mathrm{Cu}_{2} \mathrm{Cl}_{2}$.
