The element that shows greater ability to form
Question: The element that shows greater ability to form $p \pi-p \pi$ multiple bonds, is : SiGe$\mathrm{Sn}$CCorrect Option: , 4 Solution: carbon atom have $2 p$ orbitals able to form strongest $p \pi-p \pi$ bonds...
Read More →For a reaction consider the plot of
Question: For a reaction consider the plot of $\ln \mathrm{k}$ versus 1/T given in the figure. If the rate constant of this reaction at $400 \mathrm{~K}$ is $10^{-5} \mathrm{~s}^{-1}$, then the rate constant at $500 \mathrm{~K}$ is : $2 \times 10^{-4} \mathrm{~s}^{-1}$$10-4 s^{-1}$$10^{-6} \mathrm{~s}^{-1}$$4 \times 10^{-4} \mathrm{~s}^{-1}$Correct Option: , 2 Solution:...
Read More →The one that will show optical activity is :
Question: The one that will show optical activity is : (en = ethane-1,2-diamine)Correct Option: 3 Solution:...
Read More →Among the following,
Question: Among the following, the molecule expected to be stabilized by anion formation is : $\mathrm{C}_{2}, \mathrm{O}_{2}, \mathrm{NO}, \mathrm{F}_{2}$$\mathrm{NO}$$\mathrm{C}_{2}$$F_{2}$$\mathrm{O}_{2}$Correct Option: , 2 Solution: In case of only $\mathrm{C}_{2}$, incoming electron will enter in the bonding molecular orbital which increases the bond order and stability too. Whereas rest of all takes electron in their antibonding molecular orbital which decreases bond order and stability....
Read More →Solve this following
Question: The volume strength of $1 \mathrm{M} \mathrm{H}_{2} \mathrm{O}_{2}$ is: $\left(\right.$ Molar mass of $\mathrm{H}_{2} \mathrm{O}_{2}=34 \mathrm{~g} \mathrm{~mol}{ }^{-1}$ ) $16.8$$11.35$$22.4$$5.6$Correct Option: , 2 Solution: $1 \mathrm{~L}-1 \mathrm{M} \mathrm{H}_{2} \mathrm{O}_{2}$ solution will produce $11.35$ $\mathrm{L} \mathrm{O}_{2}$ gas at $\mathrm{STP}$....
Read More →Excessive release of CO_2 into the atomosphere results in :
Question: Excessive release of $\mathrm{CO}_{2}$ into the atomosphere results in :polar vortexdepletion of ozoneformation of smogglobal warmingCorrect Option: , 4 Solution: Excessive release of $\mathrm{CO}_{2}$ into the atmosphere results in global warming....
Read More →Which of the following statements is not true
Question: Which of the following statements is not true about sucrose?On hydrolysis, it produces glucose and fructoseThe glycosidic linkage is present between $C_{1}$ of $\alpha-g$ lucose and $C_{1}$ of $\beta$-fructoseIt is also named as invert sugarIt is a non reducing sugarCorrect Option: 1 Solution: Sucrose $\stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \alpha$-D-glucose $+\beta$-D-fructose also named as invert sugar \ it is a example of non-reducing sugar. The glycosidic linkage is ...
Read More →The organic compound that gives following
Question: The organic compound that gives following qualitative analysis is : Correct Option: 1 Solution:...
Read More →The combination of plots which does not represent isothermal expansion of an ideal gas is:
Question: The combination of plots which does not represent isothermal expansion of an ideal gas is: (A) and (C)(A) and (D)(B) and (D)(B) and (C)Correct Option: , 3 Solution:...
Read More →The mojor product of the following reaction is :
Question: The mojor product of the following reaction is : Correct Option: , 4 Solution:...
Read More →The degenerate orbitals of
Question: The degenerate orbitals of $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ are :$\mathrm{d}_{\mathrm{yz}}$ and $\mathrm{d}_{\mathrm{z}^{2}}$$\mathrm{d}_{z^{2}}$ and $\mathrm{d}_{\mathrm{xz}}$$d_{x z}$ and $d_{y z}$$d_{x^{2}-y^{2}}$ and $d_{x y}$Correct Option: , 3 Solution: Degenerate orbitals of $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ Hence according to the options given, degenerate orbitals are $\mathrm{d}_{\mathrm{xz}} \ \m...
Read More →The element that does NOT show catenation is:
Question: The element that does NOT show catenation is:$\mathrm{Sn}$Ge$\mathrm{Si}$$\mathrm{Pb}$Correct Option: , 4 Solution: Catenation is not shown by lead....
Read More →The correct order of the oxidation states of nitrogen in
Question: The correct order of the oxidation states of nitrogen in $\mathrm{NO}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}$ and $\mathrm{N}_{2} \mathrm{O}_{3}$ is :$\mathrm{NO}_{2}\mathrm{N}_{2} \mathrm{O}_{3}\mathrm{NO}\mathrm{N}_{2} \mathrm{O}$$\mathrm{NO}_{2}\mathrm{NO}\mathrm{N}_{2} \mathrm{O}_{3}\mathrm{N}_{2} \mathrm{O}$$\mathrm{N}_{2} \mathrm{O}\mathrm{N}_{2} \mathrm{O}_{3}\mathrm{NO}\mathrm{NO}_{2}$$\mathrm{N}_{2} \mathrm{O}\mathrm{NO}\mathrm{N}_{2} \mathrm{O}_{3}\mathrm{NO}_{2}$Correct...
Read More →The correct order of atomic radii is :
Question: The correct order of atomic radii is :$\mathrm{Ce}\mathrm{Eu}\mathrm{Ho}\mathrm{N}$$\mathrm{N}\mathrm{Ce}\mathrm{Eu}\mathrm{Ho}$$\mathrm{Eu}\mathrm{Ce}\mathrm{Ho}\mathrm{N}$$\mathrm{Ho}\mathrm{N}\mathrm{Eu}\mathrm{Ce}$Correct Option: , 3 Solution:...
Read More →The pair that does NOT require calcination is:
Question: The pair that does NOT require calcination is:$\mathrm{ZnO}$ and $\mathrm{MgO}$$\mathrm{Fe}_{2} \mathrm{O}_{3}$ and $\mathrm{CaCO}_{3} \cdot \mathrm{MgCO}_{3}$$\mathrm{ZnO}$ and $\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot \mathrm{xH}_{2} \mathrm{O}$$\mathrm{ZnCO}_{3}$ and $\mathrm{CaO}$Correct Option: 1 Solution: $\mathrm{ZnO} \ \mathrm{MgO}$ both are in oxide form therefore no change on calcination....
Read More →Among the following, the false statement is:
Question: Among the following, the false statement is:Latex is a colloidal solution of rubber particles which are positively chargedTyndall effect can be used to distingush between a colloidal solution and a true solution.It is possible to cause artificial rain by throwing electrified sand carrying charge opposite to the one on clouds from an aeroplane.Lyophilic sol can be coagulated by adding an electrolyte.Correct Option: 1 Solution: Colloidal solution fo rubber are negatively charged....
Read More →For the solution of the gases
Question: For the solution of the gases $w, x, y$ and $z$ in water at $298 \mathrm{~K}$, the Henrys law constants $\left(\mathrm{K}_{\mathrm{H}}\right)$ are $0.5,2,35$ and $40 \mathrm{kbar}$, respectively. The correct plot for the given data is :-Correct Option: , 3 Solution:...
Read More →The strength of
Question: The strength of $11.2$ volume solution of $\mathrm{H}_{2} \mathrm{O}_{2}$ is : [Given that molar mass of $\mathrm{H}=1 \mathrm{~g} \mathrm{~mol}^{-1}$ and $\mathrm{O}=16 \mathrm{~g} \mathrm{~mol}^{-1}$ ]$13.6 \%$$3.4 \%$$34 \%$$1.7 \%$Correct Option: , 2 Solution:...
Read More →The Mond process is used for the
Question: The Mond process is used for theextraction of MoPurification of $\mathrm{Ni}$Purification of $\mathrm{Zr}$ and $\mathrm{Ti}$Extraction of $\mathrm{Zn}$Correct Option: , 2 Solution: Mond's process is used for the purification of Nickel....
Read More →Solve this following
Question: If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ is $8 \times 10^{-12}$, the molar solubility of $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ in $0.1 \mathrm{M} \mathrm{AgNO} 3$ is :$8 \times 10^{-12} \mathrm{M}$$8 \times 10^{-10} \mathrm{M}$$8 \times 10^{-11} \mathrm{M}$$8 \times 10^{-13} \mathrm{M}$Correct Option: , 2 Solution:...
Read More →Polysubstitution is a major drawback in:
Question: Polysubstitution is a major drawback in:Reimer Tiemann reactionFriedel Craft's acylationFriedel Craft's alkylationAcetylation of anilineCorrect Option: , 3 Solution: In Friedal crafts alkylation product obtained is more activated and hence polysubtitution will take place....
Read More →Consider the bcc unit cells of the solids 1 and 2
Question: Consider the bcc unit cells of the solids 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is $50 \%$ more in solid 2 than in 1. What is the approximate packing efficiency in solid 2? $45 \%$$65 \%$$90 \%$$75 \%$Correct Option: , 3 Solution:...
Read More →The major product in the following conversion is :
Question: The major product in the following conversion is : Correct Option: , 2 Solution:...
Read More →If p is the momentum of the fastest electron ejected
Question: If $p$ is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength $\lambda$, then for $1.5 \mathrm{p}$ momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)$\frac{1}{2} \lambda$$\frac{3}{4} \lambda$$\frac{2}{3} \lambda$$\frac{4}{9} \lambda$Correct Option: , 4 Solution:...
Read More →Fructose and glucose can be distinguished by :
Question: Fructose and glucose can be distinguished by :Fehling's testBarfoed's testBenedict's testSeliwanoff's testCorrect Option: , 4 Solution: Seliwanoff's test is used to distinguished aldose and ketose group....
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