Solve the following
Question: Why are Mn2+compounds more stable than Fe2+towards oxidation to their +3 state? Solution: Electronic configuration of Mn2+is [Ar]183d5. Electronic configuration of Fe2+is [Ar]183d6. It isknown that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stabled5configuration. This is the reason Mn2+shows resistance to oxidation to Mn3+. Also, Fe2+has 3d6configuration and by losing one electron, its configuration changes to a more stable 3d5configuration...
Read More →Write down the electronic configuration of:
Question: Write down the electronic configuration of: (i)Cr3++(iii)Cu+(v)Co2+(vii)Mn2+ (ii)Pm3+(iv)Ce4+(vi)Lu2+(viii)Th4+ Solution: (i)Cr3+: 1s22s22p63s23p63d3 Or, [Ar]183d3 (ii)Pm3+: 1s22s22p63s23p63d104s24p64d105s25p64f4 Or, [Xe]543d3 (iii)Cu+: 1s22s22p63s23p63d10 Or, [Ar]183d10 (iv)Ce4+: 1s22s22p63s23p63d104s24p64d105s25p6 Or, [Xe]54 (v)Co2+: 1s22s22p63s23p63d7 Or, [Ar]183d7 (vi)Lu2+: 1s22s22p63s23p63d104s24p64d105s25p64f145d1 Or, [Xe]542f143d3 (vii)Mn2+: 1s22s22p63s23p63d5 Or, [Ar]183d5 (vii...
Read More →Two vertical poles of height 9 m and 14 m stand on a plane ground.
Question: Two vertical poles of height 9 m and 14 m stand on a plane ground. If the distance between their bases is 12 m, find the distance between their tops. Solution: Let the two poles be DE and AB and the distance between their bases be BE.We have:DE = 9 m, AB = 14 m and BE = 12 mDraw a line parallel to BE from D, meeting AB at C.Then, DC = 12 m and AC = 5 mWe need to find AD, the distance between their tops. Applying Pythagoras theorem in right-angled triangle ACD, we have: $A D^{2}=A C^{...
Read More →Actinoid contraction is greater from element to element than lanthanoid contraction.
Question: Actinoid contraction is greater from element to element than lanthanoid contraction. Why? Solution: In actinoids, 5forbitals are filled. These 5forbitals have a poorer shielding effect than 4forbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more that that experienced by lanthanoids. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids....
Read More →Solve the following
Question: Explain why Cu+ion is not stable in aqueous solutions? Solution: In an aqueous medium, Cu2+is more stable than Cu+. This is because although energy is required to remove one electron from Cu+to Cu2+, high hydration energy of Cu2+ compensates for it. Therefore, Cu+ion in an aqueous solution is unstable. It disproportionates to give Cu2+and Cu. $2 \mathrm{Cu}_{(a q)}^{+} \longrightarrow \mathrm{Cu}_{(a q)}^{2+}+\mathrm{Cu}_{(s)}$...
Read More →A ladder is placed in such a way that its foot is at a distance of 15 m from a wall and its top reaches a window 20 m above the ground.
Question: A ladder is placed in such a way that its foot is at a distance of 15 m from a wall and its top reaches a window 20 m above the ground. Find the length of the ladder. Solution: Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, respectively.We have:AB = 20m and BC = 15 mApplying Pythagoras theorem in right-angled triangle ABC, we get: $A C^{2}=A B^{2}+B C^{2}$ $\Rightarrow A C=\sqrt{20^{2}+15^{2}}$ $=\sqrt{400+225}$ $=\sq...
Read More →Calculate the ‘spin only’ magnetic moment of
Question: Calculate the spin only magnetic moment of M2+(aq)ion (Z= 27). Solution: Z = 27 [Ar] 3d74s2 M2+= [Ar] 3d7 3d7= i.e., 3 unpaired electrons n= 3 $\Rightarrow \sqrt{n(n+2)}=\mu$ $\Rightarrow \sqrt{3(3+2)}=\mu$ $\Rightarrow \sqrt{15}=\mu$ 4 BM...
Read More →A car travels 1 kilometre distance in which each
Question: A car travels 1 kilometre distance in which each wheel makes 450 complete revolutions. Find the radius of the its wheels. Solution: Let the radius of wheel ber. Thus, circumferenceCof the wheel $C=2 \pi r$ Since car travels 1 km distance in which wheel makes 450 complete revolutions. Then The distance covered by wheel in one revolution $=\frac{\text { Distance moved }}{\text { Number of revolution }}$ $=\frac{1000 \mathrm{~m}}{450}$ $=\frac{20}{9} \mathrm{~m}$ We know that, The distanc...
Read More →Which is a stronger reducing agent
Question: Which is a stronger reducing agent Cr2+or Fe2+and why? Solution: The following reactions are involved when Cr2+and Fe2+act as reducing agents. $\mathrm{Cr}^{2+} \longrightarrow \mathrm{Cr}^{3+} \mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}$ The $E_{\mathrm{Cr}^{2} / \mathrm{Cr}^{2}}^{\circ}$ value is $-0.41 \mathrm{~V}$ and $E_{\mathrm{Fe}^{2} / \mathrm{Fe}^{2}}^{\circ}$ is $+0.77 \mathrm{~V}$. This means that $\mathrm{Cr}^{2+}$ can be easily oxidized to $\mathrm{Cr}^{3+}$, but $\m...
Read More →A 13 m long ladder reaches a window of a building 12 m above the ground.
Question: A 13 m long ladder reaches a window of a building 12 m above the ground. Determine the distance of the foot of the ladder from the building. Solution: Let AB and AC be the ladder and height of the building.It is given that:AB = 13mand AC = 12 mWe need to find the distance of the foot of the ladder from the building, i.e, BC.In right-angled triangle ABC, we have: $A B^{2}=A C^{2}+B C^{2}$ $\Rightarrow B C=\sqrt{13^{2}-12^{2}}$ $=\sqrt{169-144}$ $=\sqrt{25}$ $=5 \mathrm{~m}$ Hence, the d...
Read More →The radii of two circles are 19 cm and 9 cm respectively.
Question: The radii of two circles are 19 cm and 9 cm respectively. Find the radius and area of the circles which has it circumference equal to the sum of the circumferences of the two circles. Solution: Let the radius of circles be $r \mathrm{~cm}, r_{1} \mathrm{~cm}$ and $r_{2} \mathrm{~cm}$ respectively. Then their circumferences are $C=2 \pi r \mathrm{~cm}, C_{1}=2 \pi r_{1} \mathrm{~cm}$ and $C_{2}=2 \pi r_{2} \mathrm{~cm}$ respectively. It is given that, Circumference $C$ of circle = Circu...
Read More →Why is the highest oxidation state of
Question: Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? Solution: Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state....
Read More →How would you account for the irregular variation of ionization enthalpies
Question: How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements? Solution: Ionization enthalpies are found to increase in the given series due to a continuous filling of the innerd-orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such asd0,d5,d10. Since these states are exceptionally stable, their ionization enthalpies are very high. In case...
Read More →A man goes 10 m due south and then 24 due west.
Question: A man goes 10 m due south and then 24 due west. How far is he from the starting point? Solution: Let the man starts from point D and goes 10 m due south and stops at E. He then goes 24 m due west at F. In right $\triangle \mathrm{DEF}$, we have: DE = 10 m, EF = 24 m $D F^{2}=E F^{2}+D E^{2}$ $D F=\sqrt{10^{2}+24^{2}}$ $=\sqrt{100+576}$ $=\sqrt{676}$ $=26 \mathrm{~m}$ Hence, the man is 26 m away from the starting point....
Read More →Solve the following
Question: The $E^{\theta}\left(\mathrm{M}^{2+} / \mathrm{M}\right)$ value for copper is positive $(+0.34 \mathrm{~V})$. What is possibly the reason for this? (Hint: consider its high $\Delta_{\mathrm{a}} H^{\theta}$ and low $\Delta_{\text {hyd }} H^{\theta}$ ) Solution: TheE(M2+/M) value of a metal depends on the energy changes involved in the following: 1.Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state. $\mathrm{M}_{(s)} \longrightar...
Read More →The radii of two circles are 19 cm and 9 cm respectively.
Question: The radii of two circles are 19 cm and 9 cm respectively. Find the radius and area of the circles which has it circumference equal to the sum of the circumferences of the two circles. Solution: Let the radius of circles be $r \mathrm{~cm}, r_{1} \mathrm{~cm}$ and $r_{2} \mathrm{~cm}$ respectively. Then their circumferences are $C=2 \pi r \mathrm{~cm}, C_{1}=2 \pi r_{1} \mathrm{~cm}$ and $C_{2}=2 \pi r_{2} \mathrm{~cm}$ respectively. It is given that, Circumference $C$ of circle $=$ Cir...
Read More →Which of the 3d series of the transition metals exhibits
Question: Which of the 3dseries of the transition metals exhibits the largest number of oxidation states and why? Solution: Mn (Z = 25) = 3d54s2 Mn has the maximum number of unpaired electrons present in thed-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7....
Read More →A man goes 80 m due east and then 150 m due north.
Question: A man goes 80 m due east and then 150 m due north. How far is he from the starting point? Solution: Let the man starts from point A and goes 80 m due east to B.Then, from B, he goes 150 m due north to C. We need to find AC.In right-angled triangle ABC, we have: $A C^{2}=A B^{2}+B C^{2}$ $A C=\sqrt{80^{2}+150^{2}}$ $=\sqrt{6400+22500}$ $=\sqrt{28900}$ $=170 \mathrm{~m}$ Hence, the man is 170 m away from the starting point....
Read More →In the series Sc (Z = 21) to Zn (Z = 30),
Question: In the series Sc (Z= 21) to Zn (Z= 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol1. Why? Solution: The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: 3d104s2), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence...
Read More →Silver atom has completely filled d orbitals
Question: Silver atom has completely filleddorbitals (4d10) in its ground state. How can you say that it is a transition element? Solution: Ag has a completely filled 4dorbital (4d105s1) in its ground state. Now, silver displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from thes-orboital. However, in the +2 oxidation state, an electron is removed from thed-orbital. Thus, thed-orbital now becomes incomplete (4d9). Hence, it is a transition element....
Read More →The radii of two circles are 8 cm and 6 cm respectively.
Question: The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles. Solution: Let the radius of circles be $r \mathrm{~cm}, r_{1} \mathrm{~cm}$ and $r_{2} \mathrm{~cm}$ respectively. Then their areas are $A=\pi r^{2} \mathrm{~cm}^{2}, A_{1}=\pi r_{1}^{2} \mathrm{~cm}^{2}$ and $A_{2}=\pi r_{2}^{2} \mathrm{~cm}^{2}$ respectively. It is given that, Area $A$ of circle $=$ Area $A_{1}$ of circle $+$ Area $A...
Read More →List the uses of Neon and argon gases.
Question: List the uses of Neon and argon gases. Solution: Uses of neon gas: (i)It is mixed with helium to protect electrical equipments from high voltage. (ii)It is filled in discharge tubes with characteristic colours. (iii)It is used in beacon lights. Uses of Argon gas: (i)Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than N. (ii)It is usually used to provide an inert temperature in a high metallurgical process. (iii)It is also used in labora...
Read More →Why do noble gases have comparatively large atomic sizes?
Question: Why do noble gases have comparatively large atomic sizes? Solution: Noble gases do not form molecules. In case of noble gases, the atomic radii corresponds to van der Waals radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, van der Waals radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period....
Read More →Why do noble gases have comparatively large atomic sizes?
Question: Why do noble gases have comparatively large atomic sizes? Solution: Noble gases do not form molecules. In case of noble gases, the atomic radii corresponds to van der Waals radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, van der Waals radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period....
Read More →The sides of certain triangles are given below. Determine which of them are right triangles.
Question: The sides of certain triangles are given below. Determine which of them are right triangles.(i) 9 cm, 16 cm, 18 cm(ii) 7 cm, 24 cm, 25 cm(iii) 1.4 cm, 4.8 cm, 5 cm(iv) 1.6 cm, 3.8 cm, 4 cm (v) $(a-1) \mathrm{cm}, 2 \sqrt{a} \mathrm{~cm},(a+1) \mathrm{cm}$ Solution: For the given triangle to be right-angled, the sum of square of the two sides must be equal to the square of the third side.Here, let the three sides of the triangle bea,bandc.(i)a= 9 cm,b= 16 cm andc= 18 cmThen, $a^{2}+b^{2...
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