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Engineering
100 Views   |

The number of octahedral void (s) per atom present in a cubic close-packed structure is:

• Option 1)

1

• Option 2)

3

• Option 3)

2

• Option 4)

4

As learnt in Voids - Voids are empty spaces in a unit cell. They are of four types: triangular, tetrahedral, octahedral, cubic. -    We know that the no. of actohedral voids in a close packed structure = no. of atoms for a cubic close packed structure, no. of atoms = 1  octahedral voids = 1   Option 1) 1 this option is correct Option 2) 3 this option is incorrect Option 3) 2 this option is...
Engineering
147 Views   |

A metal crystallizes with a face-centered cubic lattice. The edge length of the unit cell is 408 pm. The diameter of the metal atom is:

• Option 1)

288 pm

• Option 2)

408 pm

• Option 3)

144 pm

• Option 4)

204 pm

For an FCC, where r= radius of atom a= edge length Option 1) 288 pm This option is correct Option 2) 408 pm This option is incorrect Option 3) 144 pm This option is incorrect Option 4) 204 pm This option is incorrect
Engineering
92 Views   |

Equal volumes of two monoatomic gases, A and B, at same temperature and pressure are mixed. The ratio of specific heats $\left ( C_{p/C_{v}} \right )$ of the mixture will be:

• Option 1)

0.83

• Option 2)

1.50

• Option 3)

3.3

• Option 4)

1.67

Work done in adiabatic process -   - wherein     Option 1) 0.83 This option is incorrect  Option 2) 1.50 This option is incorrect  Option 3) 3.3 This option is incorrect  Option 4) 1.67 This option is correct
Engineering
99 Views   |

The orbital angular momentum of a p-electron is given as:

• Option 1)

$\frac{h}{\sqrt 2 \pi}$

• Option 2)

$\sqrt 3 \frac{h}{ 2 \pi}$

• Option 3)

$\sqrt {\frac{3}{2}}\frac{h}{\pi}$

• Option 4)

$\sqrt 6. \frac{h}{2 \pi}$

As learnt in Bohr's model - 1. Force of attraction between the nucleus and an electron is equal to centripetal force. 2.      n = principal quantum number. 3. Energy can be absorbed or emitted when electron transfer orbit   -    The orbital angular momentum is given by  where l is the azimuthal quantum number for p orbital,   orbital angular momentum =                                 ...
Engineering
125 Views   |

Which of the following statements is not valid for oxoacids of phosphorus?

• Option 1)

Orthophosphoric acid is used in the manufacture of triple superphosphate.

• Option 2)

Hypophosphorous acid is a diprotic superphosphate.

• Option 3)

All oxoacids contain tetrahedral four coordinated phosphorus.

• Option 4)

All oxoacids contain atleast one P = O and one P - OH group.

As we learnt in  Hypophosphorus Acid - H3PO2 (+1), monobasic - wherein    Hypophosphorus acid is monopratic acid with structure             Option 1) Orthophosphoric acid is used in the manufacture of triple superphosphate. This option is incorrect. Option 2) Hypophosphorous acid is a diprotic superphosphate. This option is correct. Option 3) All oxoacids contain tetrahedral four coordinated...
Engineering
95 Views   |

Standard reduction potential of the half reaction are given below

$F_{2}\left ( g \right )+2e^{-}\rightarrow 2F^{-}\left ( aq \right );E^{0}= +2.85V$

$Cl_{2}\left ( g \right )+2e^{-}\rightarrow 2Cl^{-}\left ( aq \right );E^{0}= +1.36V$

$Br_{2}\left ( 1 \right )+2e^{-}\rightarrow 2Br^{-}\left ( aq \right );E^{0}= +1.06V$

$I_{2}\left ( s \right )+2e^{-}\rightarrow 2I^{-}\left ( aq \right );E^{0}= +0.53V$

The strongest oxidising and reducing agents respectively are:

• Option 1)

$F_{2}\ and\ I^{-}$

• Option 2)

$Br_{2}\ and\ CI^{-}$

• Option 3)

$Cl_{2}\ and\ Br^{-}$

• Option 4)

$Cl_{2}\ and\ I_2$

Standard Electrode Potential - If concentration of each species taking part in the electrode reaction is unity and further the reaction is carried out at 298 K, then the potential of each electrode is said to be the standard electrode potential. -        Standard Electrode Potential - If concentration of each species taking part in the electrode reaction is unity and further the reaction is...
Engineering
114 Views   |

Sulphur trioxide can be obtained by which of the following reaction:

• Option 1)

$CaSO_{4}+C\overset{\Delta }{\rightarrow}$

• Option 2)

$Fe_{2}(SO_{4})_{3}\overset{\Delta }{\rightarrow}$

• Option 3)

$S+H_{2}SO_{4}\overset{\Delta }{\rightarrow}$

• Option 4)

$H_{2}SO_{4}+PCI_{5}\overset{\Delta }{\rightarrow}$

As learnt Sulfur trioxides - In the gaseous state monomeric So3 has a planar structure with  S-O bond distance of 143 pm and O-S-O bond angle of 120 degree. So3 molecule is a resonance hybrid - wherein     is obtained by heating Option 1) This option is incorrect Option 2) This option is correct Option 3) This option is incorrect Option 4) This option is incorrect
Engineering
97 Views   |

In which of the following arrangements the given sequence is not strictly according to the property indicated against it?

• Option 1)

$HF< HCI< HBr< HI$ : increasing acidic strength

• Option 2)

$H_{2}O< H_{2}S< H_{2}Se< H_{2}Te$ : increasing $pK_{a}$ values

• Option 3)

$NH_{3}< PH_{3}< AsH_{3}< SbH_{3}$ : increasing acidic character

• Option 4)

$CO_{2}< SiO_{2}< SnO_{2}< PbO_{2}$ : increasing oxidising power

Reducing power and acidic nature of hydrides of oxygen family - Follows the following trend  H2O -     Acidic nature of hydrides of oxygen family: The higher the acidic nature, higher the  value and  leuur will  the  value   value trend. Option 1)   : increasing acidic strength Incorrect Option 2)  : increasing  values Correct Option 3) : increasing acidic character  Incorrect Option 4) ...
Engineering
124 Views   |

The major product is:

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we learnt in  Addition of water - In presence of dilute H2SO4 alkene give addition of water according to Markovnikov's rules. - wherein     Option 1) this option is correct Option 2)   this option is incorrect Option 3) this option is incorrect Option 4) this option is incorrect
Engineering
135 Views   |

Which of the statements is not true?

• Option 1)

On passing $H_{2}S$ through acidified $K_{2}Cr_{2}O_{7}$ solution, a milky colour is observed.

• Option 2)

$Na_{2}Cr_{2}O_{7}$ is preferred over$K_{2}Cr_{2}O_{7}$ in volumetric analysis.

• Option 3)

$K_{2}Cr_{2}O_{7}$ solution in acidic medium is orange.

• Option 4)

$K_{2}Cr_{2}O_{7}$ solution becomes yellow on increasing pH beyond 7.

As we learned in concept Oxidation states - Transition elements have a variety of oxidation states but the common oxidation state is +2 for 3d metals.   -    Number of oxidation states displayed by d block elements = no. of electrons in d orbitals (unpaired) + no. of electrons in s orbitals Oxidation states in Oxidation states in Oxidation states in Oxidation states in   Option 1) On...
Engineering
123 Views   |

Four successive members of the first series of the tranisiton metals are listed below. For which one of them the standard potential $\left ( E^{\circ}_{M^{2+}/M} \right )$ value has a positive sign?

• Option 1)

Co (Z=27)

• Option 2)

Ni (Z=28)

• Option 3)

Cu (Z=29)

• Option 4)

Fe (Z=26)

M26 / M standard electrode potential - The thermochemical parameters related to the transformation of the solid metal atoms to M2+ ions in solution and their standard electrode potential E- shown in fig.   -    Among the transition metals given, only has a positive standad potential Option 1) Co (Z=27) This is incorrect option Option 2) Ni (Z=28) This is incorrect option Option 3) Cu...
Engineering
141 Views   |

Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour?

• Option 1)

$\left [ Ni(NH_{3})_{6} \right ]^{2+}$

• Option 2)

$\left [ Zn(NH_{3})_{6} \right ]^{2+}$

• Option 3)

$\left [ Cr(NH_{3})_{6} \right ]^{3+}$

• Option 4)

$\left [ Co(NH_{3})_{6} \right ]^{3+}$

As learnt in

Hybridisation -

sp3d2 - square bipyramidal or octahedral

d2sp3 - octahedral

dsp2 - square planar

- wherein

sp3d2 - outer complex

d2sp3 - inner complex

sp3 - $[Ni(Cl)_{4}]^{2-}$

dsp2 - $[Pt(CN)_{4}]^{2-}$

Electronic configuration of Ni2+ in [Ni (NH3)6]2+ : 3d8 4s° 4p°

NH3 is a strong field ligand. So,

Since, 2 orbitals are not vacant in Ni2+, so it will form an outer orbital complex with sp3d2 hybridization and 2 unpaired electrons giving it paramagnetic nature.

Option 1)

$\left [ Ni(NH_{3})_{6} \right ]^{2+}$

This option is correct

Option 2)

$\left [ Zn(NH_{3})_{6} \right ]^{2+}$

This option is incorrect

Option 3)

$\left [ Cr(NH_{3})_{6} \right ]^{3+}$

This option is incorrect

Option 4)

$\left [ Co(NH_{3})_{6} \right ]^{3+}$

This option is incorrect

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Engineering
104 Views   |

A proton carrying 1MeV kinetic energy is moving in a circular path of redius R in uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circle of same radius in the same field?

• Option 1)

2MeV

• Option 2)

1MeV

• Option 3)

0.5MeV

• Option 4)

4MeV

Radius of charged particle - -    r= Option 1) 2MeV This solution is incorrect  Option 2) 1MeV This solution is correct  Option 3) 0.5MeV This solution is incorrect  Option 4) 4MeV This solution is incorrect
Engineering
135 Views   |

In a coil of resistance 10$\Omega$, the induced current developed by changing magnetic flux through it, is shown in figure as a function of time. The magnitude of change in flux through the coil in weber is:

• Option 1)

8

• Option 2)

2

• Option 3)

6

• Option 4)

4

Induced Charge -   - wherein Induced Charge time independent     Amount of charge passes through the coil Area under graph=   Option 1) 8 This is incorrect option Option 2) 2 This is correct option Option 3) 6 This is incorrect option Option 4) 4 This is incorrect option
Engineering
92 Views   |

A coil of resistance 400$\Omega$ is placed in a magnetic field. If the magnetic flux $\phi$ (wb) linked with the coil varies with time t ($\sec$) as $\phi=50t^{2}+4$ . The current in the coil at t=2 sec is:

• Option 1)

0.5A

• Option 2)

0.1A

• Option 3)

2A

• Option 4)

1A

Faraday Second Law of Induction emf - -     Option 1) 0.5A This is correct option Option 2) 0.1A This is incorrect option Option 3) 2A This is incorrect option Option 4) 1A This is incorrect option
Engineering
216 Views   |

Consider the reaction:

RCHO+NH2NH2 —> RCH=N-NH2

What sort of reaction is it?

• Option 1)

• Option 2)

• Option 3)

Electrophilic substitution - elimination reaction

• Option 4)

As learnt in Chemical properties of aldehydes and ketones - Undergo nucleophilic addition reactions at carbon-oxygen double bond, hybridization of carbon changes from sp2 to  sp2.  Aldehydes are more reactive than ketones due to steric hindrance and  +I effect of an alkyl group. -    and Addition of ammonia and its derivatives with aldehydes and ketones - Product formation is favoured due to...
Engineering
106 Views   |

Consider the following reaction:

• Option 1)

C6H5CHO

• Option 2)

C6H5OH

• Option 3)

C6H5COCH3

• Option 4)

C6H5Cl

As we learnt in Rosenmund's reaction - Acid chlorides are reduced in boiling xylene using Pd / Pt along with BeSO4. S or quinoline is used to poison  Pd / Pt to stop reduction of aldehyde further. - wherein   Option 1) C6H5CHO This option is correct Option 2) C6H5OH This option is incorrect Option 3) C6H5COCH3 This option is incorrect Option 4) C6H5Cl This option is incorrect
Engineering
97 Views   |

In the following sequence of reactions

$CH_{3}-Br\overset{KCN}{\rightarrow}A\overset{H_{3}O^{+}}{\rightarrow}B\xrightarrow[ether]{LiAlH_{4}}C$ , the end product (C) is

• Option 1)

Acetone

• Option 2)

Methane

• Option 3)

Acetaldehyde

• Option 4)

Ethyl alcohol

As we learnt in  Alcohol formation by reduction of carboxylic acids - Yields primary alcohol. - wherein     Preparation of alcohol - By treatment with  ag.KOH  or moist  Ag2O - wherein       Option 1) Acetone This solution is incorrect  Option 2) Methane This solution is incorrect  Option 3) Acetaldehyde This solution is incorrect  Option 4) Ethyl alcohol This solution is correct
Engineering
365 Views   |

(A) In equisetum the female gametophyte is retained on the parent sporophyte

(B) In Ginkgo male gametophyte is not independent

(C) The sporophyte is Riccia is more developed than that in polytrichum

(D) Sexual reproduction in Volvox is isogamous

(E) The spores of slime molds lack cell walls. How many of the above statements are correct?

• Option 1)

Two

• Option 2)

Three

• Option 3)

Four

• Option 4)

one

option 4

Engineering
129 Views   |

Cycas and adiantum resemble each other in having:

• Option 1)

seeds

• Option 2)

Motile sperms

• Option 3)

cambium

• Option 4)

vessels

Pterosida / Filicopsida (Ferns) - 1-Leaves are fan likes. 2-Rhizome stem present. 3-Homosporous or heterosporus. - wherein Eg. Dryopteris, pteris, Adiantum.     Gymnosperms - Evergreen, perennial woody trees or shrubs. They are non-flowering plants producing naked seeds. - wherein Eg Cycas, Pinus, Cedrus   In both cycas and adiantum motile sperms are found Option 1) seeds  This is incorrect...
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