Write down the IUPAC name for each of the following complexes and indicate the oxidation state,
Question: Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex: (i)K[Cr(H2O)2(C2O4)2].3H2O (ii)[Co(NH3)5Cl]Cl2 (iii)CrCl3(py)3 (iv)Cs[FeCl4] (v)K4[Mn(CN)6] Solution: (i)Potassium diaquadioxalatochromate (III) trihydrate. Oxidation state of chromium = 3 Electronic configuration: 3d3:t2g3 Coordination number = 6 Shape: octahedral Stereochemistry:...
Read More →Two poles of height 13 m and 7 m respectively stand vertically on a plane ground at a distance of 8 m from each other.
Question: Two poles of height 13 m and 7 m respectively stand vertically on a plane ground at a distance of 8 m from each other. The distance between their tops is(a) 9 m(b) 10 m(c) 11 m(d) 12 m Solution: (b) 10 m Let AB and DE be the two poles.According to the question:AB = 13 mDE = 7 mDistance between their bottoms = BE = 8 mDraw a perpendicularDC to AB from D, meeting AB at C. We get:DC = 8m, AC = 6 mApplying Pythagoras theorem in right-angled triangle ACD, we have: $A D^{2}=D C^{2}+A C^{2}$ ...
Read More →A toy is in the form of a cone surmounted on a hemisphere.
Question: A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm, respectively. Determine the surface area of the toy. (Use = 3.14) Solution: Given that, a toy is in the form of a cone surmounted on the hemisphere. Diameter of the base $d=6 \mathrm{~cm}$ and the height of the cone $h=4 \mathrm{~cm}$, then we have to find the surface area of the toy. We have the following figure The radius of the base is $r=\frac{d}{2}$ $=\...
Read More →A man goes 24 m due west and them 10 m due north. How far is he from the starting point?
Question: A man goes 24 m due west and them 10 m due north. How far is he from the starting point?(a) 34 m(b) 17 m(c) 26 m(d) 28 m Solution: (c) 26 m Suppose, the man starts from point A and goes 24 m due west to point B. From here, he goes 10 m due north and stops at C.In right triangle ABC, we have:AB = 24 m, BC = 10 mApplying Pythagoras theorem, we get: $A C^{2}=A B^{2}+B C^{2}=24^{2}+10^{2}$ $A C^{2}=576+100=676$ $A C=\sqrt{676}=26$...
Read More →Give the oxidation state,
Question: Give the oxidation state,d-orbital occupation and coordination number of the central metal ion in the following complexes: (i)K3[Co(C2O4)3] (ii)cis-[Cr(en)2Cl2]Cl (iii)(NH4)2[CoF4] (iv)[Mn(H2O)6]SO4 Solution: (i)K3[Co(C2O4)3] The central metal ion is Co. Its coordination number is 6. The oxidation state can be given as: x 6 = 3 x= + 3 Thedorbital occupation for Co3+ist2g6eg0. (ii)cis-[Cr(en)2Cl2]Cl The central metal ion is Cr. The coordination number is 6. The oxidation state can be gi...
Read More →A tent of height 77 dm is in the form of a right circular
Question: A tent of height $77 \mathrm{dm}$ is in the form of a right circular cylinder of diameter $36 \mathrm{~m}$ and height $44 \mathrm{dm}$ surmounted by a right circular cone. Find the cost of the canvas at Rs. $3.50$ per $m^{2}$ (Use $\pi=22 / 7$ ). Solution: Given: Height of the tent $\mathrm{h}=77 \mathrm{dm}=7.7 \mathrm{~m}$, diameter of cylinder $d=36 \mathrm{~m}$ Height of the cylinder h1= 44 dm = 4.4 m, height of cone h2= 33 dm = 3.3 m We have the following diagram Radius $r=\frac{d...
Read More →Discuss the nature of bonding in metal carbonyls.
Question: Discuss the nature of bonding in metal carbonyls. Solution: The metal-carbon bonds in metal carbonyls have both and characters. A bond is formed when the carbonyl carbon donates a lone pair of electrons to the vacant orbital of the metal. A bond is formed by the donation of a pair of electrons from the filled metaldorbital into the vacant anti-bonding * orbital (also known as back bonding of the carbonyl group). The bond strengthens the bond and vice-versa. Thus, a synergic effect is c...
Read More →Solve the following
Question: [Fe(CN)6]4and [Fe(H2O)6]2+are of different colours in dilute solutions. Why? Solution: The colour of a particular coordination compound depends on the magnitude of the crystal-field splitting energy, Δ. This CFSE in turn depends on the nature of the ligand. In case of [Fe(CN)6]4and [Fe(H2O)6]2+, the colour differs because there is a difference in the CFSE. Now, CNis a strong field ligand having a higher CFSE value as compared to the CFSE value of water. This means that the absorption o...
Read More →A solution of
Question: A solution of [Ni(H2O)6]2+is green but a solution of [Ni(CN)4]2is colourless. Explain. Solution: In $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}, \mathrm{H}_{2} \ddot{\mathrm{O}}$ is a weak field ligand. Therefore, there are unpaired electrons in $\mathrm{Ni}^{2+} .$ In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of $d-d$ transition is present. $\left.\operatorname{Hence}, \mathrm{...
Read More →A rocket is in the form of a circular cylinder closed
Question: A rocket is in the form of a circular cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of radius 2.5 m and height 21 m and the cone has the slant height 8 m. Calculate the total surface area and the volume of the rocket. Solution: Given: Radius of the cylinder $r=2.5 \mathrm{~m}$, height of the cylinder, $h=21 \mathrm{~m}$, slant height of the cone $l=8 \mathrm{~m}$. We have to find total surface area and volume of the rocket Let us a...
Read More →Explain why?
Question: [Cr(NH3)6]3+is paramagnetic while [Ni(CN)4]2is diamagnetic. Explain why? Solution: Cr is in the +3 oxidation state i.e.,d3configuration. Also, NH3is a weak field ligand that does not cause the pairing of the electrons in the 3dorbital. Cr3+ Therefore, it undergoesd2sp3hybridization and the electrons in the 3dorbitals remain unpaired. Hence, it is paramagnetic in nature. In [Ni(CN)4]2, Ni exists in the +2 oxidation state i.e.,d8configuration. Ni2+: CNis a strong field ligand. It causes ...
Read More →For each of the following statements state whether true(T) or false(F).
Question: For each of the following statements state whether true(T) or false(F).(i) Two circles with different radii are similar.(ii) Any two rectangles are similar.(iii) If two traingles are similar, the their corresponding angles are equal and their corresponding sides are equal.(iv) The length of the line segment joining the mid points of any two sides of a trinagle is equal to the half the length of the third side.(v) In △ABC, AB = 6 cm, A = 450and AC = 8 cm and in △DEF, DF = 9 cm, D = 450a...
Read More →A tent is in the form of a right circular cylinder surmounted by a cone.
Question: A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of canvas required for the tent. Solution: We have a right circular cylinder surmounted by a cone. Diameter of cylinder = 24 m, Height if cylindrical portion = 11 m and the vertex of the cone is 16 meters above the ground. We have to find the area of canvas required f...
Read More →What is crystal field splitting energy?
Question: What is crystal field splitting energy? How does the magnitude of Δodecide the actual configuration ofd-orbitals in a coordination entity? Solution: The degenerated-orbitals (in a spherical field environment) split into two levels i.e.,egandt2gin the presence of ligands. The splitting of the degenerate levels due to the presence of ligands is called the crystal-field splitting while the energy difference between the two levels (egandt2g) is called the crystal-field splitting energy. It...
Read More →What is spectrochemical series?
Question: What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand. Solution: A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy (CFSE) values. The ligands present on the R.H.S of the series are strong field ligands while that on the L.H.S are weak field ligands. Also, strong field ligands cause higher splitting in thedorbitals than weak field ligands. I Br S2 SCN Cl N3 ...
Read More →The difference between the outer and inner
Question: The difference between the outer and inner curved surface areas of a hollow right circular cylinder 14 cm long is 88 cm2. If the volume of metal used in making the cylinder is 176 cm3, find the outer and inner diameters of the cylinder. (Use = 22/7) Solution: The height of the hollow cylinder is 14 cm. Let the inner and outer radii of the hollow cylinder arercm andRcm respectively. The difference between the outer and inner surface area of the hollow cylinder is $=2 \pi R \times 14-2 \...
Read More →Draw figure to show the splitting of d orbitals in an octahedral crystal field.
Question: Draw figure to show the splitting ofdorbitals in an octahedral crystal field. Solution: The splitting of the $d$ orbitals in an octahedral field takes palce in such a way that $d_{x^{2}-y^{2}}, d_{z^{2}}$ experience a rise in energy and form the $e_{g}$ level, while $d_{x y}, d_{y z}$ and $d_{z x}$ experience a fall in energy and form the $t_{2 g}$ level....
Read More →Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
Question: Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (i)[Fe(CN)6]4 (ii)[FeF6]3 (iii)[Co(C2O4)3]3 (iv)[CoF6]3 Solution: (i)[Fe(CN)6]4 In the above coordination complex, iron exists in the +II oxidation state. Fe2+: Electronic configuration is 3d6 Orbitals of Fe2+ion: As CNis a strong field ligand, it causes the pairing of the unpaired 3delectrons. Since there are six ligands around the central metal ion, the most feasible hybridizatio...
Read More →A cylindrical bucket, 32 cm high and with radius of base 18 cm,
Question: A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied out on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap. Solution: Let the radius of the cone byrNow, Volume cylindrical bucket = Volume of conical heap of sand $\Rightarrow \pi(18)^{2}(32)=\frac{1}{3} \pi r^{2}(24)$ $\Rightarrow(18)^{2}(32)=8 r^{2}$ $\Rightarrow r^{2}=18 \times 18 \times 4$ ...
Read More →The length of the diagonals of a rhombus are 40 cm and 42 cm.
Question: The length of the diagonals of a rhombus are 40 cm and 42 cm. Find the length of each sides of the rhombus Solution: Suppose ABCD is a rhombus.We know that the digonals of a rhombus perpendicularly bisect each other. AOB = 900, AO = 20 cm and BO = 21 cmNow, In right triangle AOBBy using Pythagoras theorem we haveAB2= AO2+ OB2= 202+ 212= 400 + 441= 841AB2= 841⇒ AB = 29 cmSince, all the sides of a rhombus are equal.Hence, AB = BC = CD = DA = 29 cm...
Read More →What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate?
Question: What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution? Solution: i.e., Thus, the coordination entity formed in the process is $\mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right] . \mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]$ is a very stable complex, which does not ionize to give $\mathrm{Cu}^{2+}$ ions...
Read More →In the given figure, ∠AMN = ∠MBC = 760 If p, q and r are the lengths
Question: In the given figure, AMN = MBC = 760Ifp,qandrare the lengths of AM, MB and BC respectively, then express the length of MN in terms ofp,qandr. Solution: In △AMN and △ABCAMN = ABC = 760 (Given)A = A (Common)By AA similarity criterion, △AMN △ABCIf two triangles are similar, then the ratio of their corresponding sides are proportional $\therefore \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{MN}}{\mathrm{BC}}$ $\Rightarrow \frac{\mathrm{AM}}{\mathrm{AM}+\mathrm{MB}}=\frac{\mathrm{MN}}{\mat...
Read More →Aqueous copper sulphate solution (blue in colour) gives:
Question: Aqueous copper sulphate solution (blue in colour) gives: (i)a green precipitate with aqueous potassium fluoride, and (ii)a bright green solution with aqueous potassium chloride Explain these experimental results. Solution: Aqueous CuSO4exists as [Cu(H2O)4]SO4. It is blue in colour due to the presence of [Cu[H2O)4]2+ions. (i)When KF is added: (ii)When KCl is added: In both these cases, the weak field ligand water is replaced by the Fand Clions....
Read More →The volume of a hemi-sphere is
Question: The volume of a hemi-sphere is $2425 \frac{1}{2} \mathrm{~cm}^{3}$. Find its curved surface area. (Use $\pi=22 / 7$ ) Solution: Let the radius of the hemisphere bercm. Volume of hemisphere $=2425 \frac{1}{2} \mathrm{~cm}^{3}$ $\Rightarrow \frac{2}{3} \pi r^{3}=\frac{4851}{2}$ $\Rightarrow \frac{2}{3} \times \frac{22}{7} r^{3}=\frac{4851}{2}$ $\Rightarrow r^{3}=\frac{4851 \times 3 \times 7}{2 \times 2 \times 22}$ $\Rightarrow r^{3}=\frac{441 \times 21}{2 \times 2 \times 2}$ $\Rightarrow...
Read More →Write all the geometrical isomers of
Question: Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers? Solution: [Pt(NH3)(Br)(Cl)(py) From the above isomers, none will exhibit optical isomers. Tetrahedral complexes rarely show optical isomerization. They do so only in the presence of unsymmetrical chelating agents....
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