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Medical
98 Views   |

Solubility of the alkaline earth's metal sulphates in water decrease in the sequence:-

• Option 1)

Ca > Sr > Ba > Mg

• Option 2)

Sr > Ca > Mg > Ba

• Option 3)

Ba > Mg > Sr > Ca

• Option 4)

Mg > Ca > Sr > Ba

As we learnt in Solubility of sulphates of alkaline metals - Solubility in water decreases from Be to Ba - wherein Lattice energy is constant whereas hydration energy decreases from    Solubility of surface of alkaline earth metals decreases down the group from Be to BA  because lattoce emergy remains almost constant, whereas the hydration energy decreases dramatically.    Option 1) Ca > Sr...
Medical
115 Views   |

The function of "Sodium pump" is a biological process operating in each and every cell of all animals. Which of the following biologically important ions is also a consituent of this pump:

• Option 1)

$Mg^{2+}$

• Option 2)

$K^{+}$

• Option 3)

$Fe^{2+}$

• Option 4)

$Ca^{2+}$

As we learnt in concept Sodium-Potassium Pump - An ATPase which operates across the plasma membrane and allows the exchange of K+ and Na+  against the concentrated gradients. -    The sodium pump (also called sodium potassium pump) is an AT Pase which operates across the plasma membrane and allows the exchange of K+ and  Na+ against the concentrated gradients. Option 1) This solution is...
Medical
117 Views   |

A gas such as carbon monoxide would be most likely to obey the ideal gas law at:

• Option 1)

High temperatures and low pressure.

• Option 2)

Low temperatures and high pressure.

• Option 3)

High temperatures and high pressure.

• Option 4)

Low temperatures and low pressure.

Ideal Gas Law - - wherein P - Pressure V - Volume n - No. of Moles R - Gas Constant T - Temperature    A real gas can behave as an ideal gas if volume of molecule is negligible and interaction between the molecule is negligible. This will happen when number of molecule per unit volume is very small.  From ideal gas equation,  will be very large if  is very large and  is very...
Medical
111 Views   |

The stability of +1 oxidation state among Al, Ga, In and Tl increases in the sequence:

• Option 1)

Ga < In < Al < Tl

• Option 2)

Al < Ga < In < Tl

• Option 3)

Tl < In < Ga < Al

• Option 4)

In < Tl < Ga < Al

As we learned in concept no Inert pair effect - The phenomenon in which outer shell s electrons (ns2) penetrate to (n-1) d-electrons and thus become closer to nucleus and are more effectively pulled by the nucleus - wherein This results in less availability of ns2 electrons pair for bonding or ns2 electron pair becomes inert.    The order of stability of +1 oscidation state is This is on...
Medical
93 Views   |

The variation of the boiling point of the hydrogen halides is in the order $HF> HI> HBr> HCl$.
What explains the higher boiling point of hydrogen fluoride?

• Option 1)

The electronegativity of fluourine is much higher than for other elements in the group

• Option 2)

There is strong hydrogen bonding between HF molecules

• Option 3)

The bond energy of HF molecules is greater than in other hydrogen halides.

• Option 4)

The effect of nuclear shielding is much reduced in fluorine which polarises the HF molecule.

As learnt in Boiling point of Hydride - HF>HI>HBr>HCl - wherein Boiling point of HF is due to H-Bond     Order of boiling point : HF > HI > HBr > HCl HF has highest boiling point because of strong hydrogen bonding. Option 1) The electronegativity of fluourine is much higher than for other elements in the group This option is incorrect Option 2) There is strong hydrogen bonding between HF...
Medical
152 Views   |

Given:

The enthalpy of the hydrogenation of these compounds will be in the order as:

• Option 1)

$III> II> I$

• Option 2)

$II> III> I$

• Option 3)

$II> I> III$

• Option 4)

$I> II> III$

As learnt Preparation of Alkane from Unsaturated Hydrocarbon - Dihydrogen gas adds to alkenes and alkynes in the presence of finely divided catalyst like Pt, Pd or Ni to form alkane. - wherein    Addition of hyrogen to a compound is known as hydrogenation. Enthalpy of hydrogenation Option 1) This option is correct Option 2) This option is incorrect Option 3) This option is...
Medical
88 Views   |

The reaction of $C_{6}H_{5}CH=CHCH_{3}$  with HBr produces:

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we learnt in  Addition of Hydrogen Halide on alkene - hydrogen halides add up to alkenes to form alkyl halides. The order of reactivity of the hydrogen halides is HI>HBr>HCl - wherein                                                                       Option 1) this option is incorrect Option 2) this option is incorrect Option 3) this option is incorrect Option 4) this option is correct
Medical
117 Views   |

Which of the following pairs of ions are isoelectronic and isostructural?

• Option 1)

$ClO^{-}_{3}, CO^{2-}_{3}$

• Option 2)

$SO^{2-}_{3}, NO^{-}_{3}$

• Option 3)

$ClO^{-}_{3}, SO^{2-}_{3}$

• Option 4)

$CO^{2-}_{3}, SO^{2-}_{3}$

As we learnt in  Isosteres - Molecules or polyatomic ions containing same number of atoms and the same total numbers of electron are called isosteres.Generally isosteres have identical lewis structure.       -   Isolectronic and Isotructural units can also be called isosteres No. of electrons in No. of electrons in  and , both are hybridized and thus have similar structure. Option 1) This...
Medical
98 Views   |

The total number of $\pi$ -bond electrons in the following structure is:

• Option 1)

8

• Option 2)

12

• Option 3)

16

• Option 4)

4

Preparation of Alkane from Unsaturated Hydrocarbon - Dihydrogen gas adds to alkenes and alkynes in the presence of finely divided catalyst like Pt, Pd or Ni to form alkane. - wherein    Unsaturated double bond hydrocarbon determines the number of  electrons= 2Xnumber of double bonds = 2x4 = 8 Option 1) 8 this option is correct Option 2) 12 this option is incorrect Option 3) 16 this option...
Medical
106 Views   |

The hybridization involved in complex $\left [ Ni \left ( CN \right )_{4} \right ]^{2-}$is.(At.No.Ni=28)

• Option 1)

$dsp^{2}$

• Option 2)

$sp^{3}$

• Option 3)

$d^{2}sp^{2}$

• Option 4)

$d^{2}sp^{3}$

As we learnt in  Hybridisation - The process of mixing of atomic orbitals belonging to the same atoms of slightly different energies so that a redistribution of energy takes place between them  resulting in the formation of new set of orbital of equivalent energies and shape is called hybridisation. - wherein The new orbitals thus formed are called hybrid orbitals    In       or we could...
Medical
188 Views   |

In which of the following pairs, both the species are not isostructural?

• Option 1)

$SiCl_{4}, PCl^{+}_{4}$

• Option 2)

Diamond, silicon carbide

• Option 3)

$NH_{3}, PH_{3}$

• Option 4)

$XeF_{4}, XeO_{4}$

As we learnt in  Dettermination of shape of molecules using VSEPR Theory - -     Shape of molecules - -     V = No. of valence electrons M = No. of surrounding monavalent atoms  C = cationic charge  A = anionic charge  In , sp3d2 hybridization. In , sp3 hybridization  is square planar, whereas XeO4 is tetrahedrol.   Option 1) This solution is incorrect  Option 2) Diamond, silicon...
Medical
140 Views   |

Which of the following processes does not involve oxidation of iron?

• Option 1)

Decolourization of blue $CuSO_{4}$ solution by iron

• Option 2)

Formation of $Fe(CO)_{5}$ from $Fe$

• Option 3)

Liberation of $H_{2}$ from steam by iron at high temperature

• Option 4)

Rusting of iron sheets

As learnt in Oxidation states - Transition elements have a variety of oxidation states but the common oxidation state is +2 for 3d metals.   -    The reaction goes as follows: Clearly, there is no change in oxidation state of iron. Option 1) Decolourization of blue  solution by iron This option is incorrect Option 2) Formation of  from  This option is correct Option 3) Liberation of  from...
Medical
141 Views   |

Which is the correct order of increasing energy of the listed orbitals in the atom of titanium?

• Option 1)

$3s\:\: 4s \:\: 3p \:\: 3d$

• Option 2)

$4s \:\: 3s \:\: 3p \:\: 3d$

• Option 3)

$3s \:\: 3p \:\: 3d \:\: 4s$

• Option 4)

$3s \:\: 3p\:\: 4s \:\: 3d$

As learnt Electronic Configuration - The 'd' block elements follow the general configuration . -    According to the Aufbau Principle, e-s occupy lower energy orbitals before going higher. Electrons are assigned to orbitals in order of increasing value of n+l. For subshells with some value of (n+l), e-s are assigned first to the sub shell with lower x.                 n+l...
Medical
93 Views   |

The number of $d$-electrons in $Fe^{2+}(Z=26)$ is not equal to the number of electrons in which one of th e following?

• Option 1)

$p$-electrons in $cl (Z=17)$

• Option 2)

$d$-electrons in $Fe (Z=26))$

• Option 3)

$p$-electrons in $Ne (Z=10)$

• Option 4)

$s$-electrons in $Mg (Z=12)$

Electronic Configuration - The 'd' block elements follow the general configuration. -    Electronic Config of      :  no of d Electronic = 6 Electronic Config Of Cl :    no of p Electronic =  Option 1) -electrons in  Correct Option 2) -electrons in  Incorrect Option 3) -electrons in  Incorrect Option 4) -electrons in  Incorrect
Medical
116 Views   |

Which of these statements about $\left [ Co(CN)_{6} \right ]^{3-}$ is true?

• Option 1)

$\left [ Co (CN_{6}) \right ]^{3-}$ has four unpaired electrons and will be in a low-spin configuration.

• Option 2)

$\left [ Co(CN)_{6} \right ]^{3-}$ has four unpaired electrons and will be in a high spin configuration.

• Option 3)

$\left [ Co(CN)_{6} \right ]^{3-}$ has no unpaired electrons and will be in a high-spin configuration.

• Option 4)

$\left [ Co(CN)_{6} \right ]^{3-}$ has no unpaired electrons and will be in a low-spin configuration.

As we learnt in  Magnetic Moments (spin only) -  where n= number of unpaired electron. - wherein Number of unpaired  and corresponding magnetic moments    Electronic config of for    = will be paired up because is an SFL has no unpaired and is in a low spin config.     Option 1) has four unpaired electrons and will be in a low-spin configuration. This solution is incorrect. Option...
Medical
132 Views   |

The sum of coordination number and oxidation number of the metal M in the complex $\left [ M(en)_{2} (C_{2}O_{4})\right ]Cl$ (where $en$ is ethyenediamine) is:

• Option 1)

9

• Option 2)

6

• Option 3)

7

• Option 4)

8

As we learned in concept Types of Ligands on the basis of Connectivity - (i) Unidentate - ligand is bound to metal ion through a single donor atom. eg:  (ii) Bidentate : when ligand can bind through two donor atoms eg:  (iii) Polydentate - when ligand bind to two or more donor atoms (iv) Hexadentate - type of polydentate having six donor atoms (v) Ambidentate- which can ligate through two...
Medical
92 Views   |

The name of complex ion, $\left [ Fe(CN)_{6} \right ]^{3-}$ is:

• Option 1)

Hexacyanoiron (III) ion

• Option 2)

Hexacyanitoferrate (III) ion

• Option 3)

Tricyanoferrate (III) ion

• Option 4)

Hexacyanidoferrate (III) ion

As we learned in concept Writing the name of Complex compound formula - (i) Cation is named first  (ii) ligands are named in alphabetical order before name of central metal atom. (iii) Prefixes, mono, di, tri are used to indicate the no of ligands (iv) when name of ligand used a numerical prefix, then term bix, tris, tetrakis are used. eg: is named as dichlorobis...
Medical
86 Views   |

Number of possible isomers for the complex $\left [ Co(en)_{2}Cl_{2} \right ]Cl$ will be (en = ethylenediamine)

• Option 1)

2

• Option 2)

1

• Option 3)

3

• Option 4)

4

As we learnt in Stereoisomers - Stereoisomers have the same chemical formula and chemical bonds but they have different special arrangement -    , possible isomers are: Since, this product is chiral, so it's mirror image will also be a possible isomer. This one is not chiral. So, no mirror image isomer.  total possible isomers =  Option 1) 2 This solution is incorrect Option 2) 1 This...
Medical
97 Views   |

Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test of chloride ions with silver nitrate at $25^{\circ}C$?

• Option 1)

$CoCl_{3}\cdot 4NH_{3}$

• Option 2)

$CoCl_{3}\cdot 5NH_{3}$

• Option 3)

$CoCl_{3}\cdot 6NH_{3}$

• Option 4)

$CoCl_{3}\cdot 3NH_{3}$

As learned in concept no Coordination Sphere - The central atom/ion and the ligands attached to metal are enclosed in square bracket and this collectively termed as coordination sphere - wherein eg:   Counter ion   Coordination sphere      The co-ordination number of Cobalt (III) in octahedral complex is 6. In , all we have is 6 ligands, which implies all of them are in the co-ordination...
Medical
111 Views   |

The formation of the oxide ion O2-(g), from oxygen atom requires first an exothermic and then an endothermic step as shown below:

$O(g)+e^{-}\rightarrow O^{-}(g); \Delta_{f}H^{\ominus}=-141kJ\:\:mol^{-1}$

$O^{-}(g)+e^{-}\rightarrow O^{2-}(g); \Delta_{f}H^{\ominus}=+780kJ\:\:mol^{-1}$

Thus process of formation of $O^{2-}$  in gas phase is unfavourable even though $O^{2-}$  is isoelectronic with neon. It is due to the fact that

• Option 1)

Electron repulsion outweighs the stability gained by achieving noble gas configuration

• Option 2)

$O^{-}$ ion has comparatively smaller size than oxygen atom.

• Option 3)

Oxygen is more electronegative.

• Option 4)

Addition of electron in oxygen results in larger size of the ion.

As learnt in Variation of electron gain enthalpy with Zeff - The higher the Zeff the greater is the tendency of the atom to attract the incoming electron towards itself and hence higher the amount of energy released. - wherein    We know that   After gaining 1 electron, the value of Zeff falls and thus gaining another electron becomes unfavourable to such an extent that the process becomes...
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