InterviewSolution

Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

51.	What is the end product formed when KMnO4 reacts with HCl?(a) Brown fumes(b) Dark green gas(c) Greenish yellow gas(d) Dense white fumesThis question was addressed to me at a job interview.Origin of the question is Application of d and f-Block Elements in division D and F-Block Elements of Chemistry – Class 12
Answer» Correct answer is (c) Greenish yellow GAS The best I can EXPLAIN: Potassium permanganate (KMnO4) REACTS with hydrochloric acid(HCl) to form potassium chloride, manganese chloride, CHLORINE and water. Here, one of the products formed, chlorine, is a greenish yellow gas. It has a pungent SMELL and is toxic.

52.	Choose the correct statement.(a) Both actinoids and lanthanoids are less basic(b) Both actinoids and lanthanoids do not show same oxidation of +3(c) Both actinoids and lanthanoids do not exhibit magnetic and spectral properties(d) Both actinoids and lanthanoids are electropositiveI got this question during an online interview.This intriguing question originated from D and F-Block Elements topic in chapter D and F-Block Elements of Chemistry – Class 12
Answer» Correct option is (d) Both actinoids and lanthanoids are electropositive Explanation: LANTHANIDES and ACTINIDES involve filling of f-orbitals and thus are similar in MANY respects. The most common OXIDATION state is +3 for both lanthanides and actinides. Both are electropositive in nature and thus very reactive. Magnetic and spectral properties are exhibited by both lanthanides and actinides.

53.	Which metal is used in an aircraft turbine engine?(a) Iron(b) Thorium(c) Titanium(d) NeodymiumThis question was addressed to me in quiz.My question is based upon Application of d and f-Block Elements topic in portion D and F-Block Elements of Chemistry – Class 12
Answer» Right choice is (c) Titanium The explanation: Titanium is one such material which PLAYS a pivotal role in modern jet engine design. Its STRENGTH to WEIGHT ratio and RESISTANCE to extreme heat make it the ideal candidate for aerospace applications. Hence, it is used in aircraft TURBINE engines.

54.	Which was the first transuraniumactinoid element to be discovered?(a) Neptunium(b) Americium(c) Protactinium(d) ActiniumThe question was asked in quiz.My question is from D and F-Block Elements topic in portion D and F-Block Elements of Chemistry – Class 12
Answer» The CORRECT CHOICE is (a) Neptunium For explanation: Neptunium was the first transuranium element to be discovered in 1940. It is a silvery metal, with a MELTING POINT of 637°C and a boiling point of 4174°C. It exhibits oxidation states of +3 to +7 in compounds and useful for chemical experiments.

55.	Which of the following is not a consequence of lanthanide contraction?(a) From La^+3 to Lu^+3, the ionic radii changes from 106 pm to 85 pm(b) As the size of the lanthanide ions decreases the basic strength increases(c) The basic character of oxides and hydroxides decreases with increase in atomic number(d) The atomic radii of 4d and 5d series is similarI had been asked this question by my school principal while I was bunking the class.Question is taken from D and F-Block Elements topic in portion D and F-Block Elements of Chemistry – Class 12
Answer» Correct answer is (b) As the SIZE of the lanthanide ions decreases the basic strength INCREASES For explanation: The small average decreases in the atomic size is responsible for a small DECREASE in ELECTRONEGATIVITY and S.O.P of lanthanides. As the size of the lanthanide ions decreases the COVALENT character of M—OH bond increases and hence basic strength decreases.

56.	Why do transition elements form alloys so easily?(a) Atomic size(b) Orbital configuration(c) Very light(d) Hard elementsI had been asked this question in quiz.Question is taken from Electronic Configurations of the D-Block Elements topic in portion D and F-Block Elements of Chemistry – Class 12
Answer» Right CHOICE is (a) Atomic size To explain: ALLOYS are combinations of metals or elements in the form of a compound or a solution. Transition elements are highly capable in forming alloys because they have similar atomic size and can SUBSTITUTE each of their POSITIONS in a crystal lattice.

57.	Which of the following is Baeyer’sreagent?(a) Acidified KMnO4(b) Alkaline KMnO4(c) Acidified K2Cr2O7(d) Aqueous KMnO4The question was posed to me by my college director while I was bunking the class.The question is from D and F-Block Elements topic in portion D and F-Block Elements of Chemistry – Class 12
Answer» Correct choice is (b) Alkaline KMnO4 Easiest explanation: Alkaline KMnO4 is called Baeyer’s reagent. Baeyer’s reagent is an alkaline solution of COLD potassium PERMANGANATE, which is a powerful oxidant making this a redox reaction. Reaction with double or triple BONDS (-C=C- or -C≡C-) in an organic material causes the colour to fade from purplish-pink to brown. It is a SYN addition reaction.

58.	Which of the following element belongs to the third transition series of the periodic table?(a) Sc(b) Re(c) Rh(d) TcThis question was addressed to me in semester exam.My enquiry is from Position of the Periodic Table in section D and F-Block Elements of Chemistry – Class 12
Answer» Right option is (b) Re Best explanation: The third transition series or 5D series corresponding to the filling of 5d sublevel consists of the following 10 elements of the 6^th period: La (Atomic No. = 57); Hf (Atomic No. = 72), Ta, W, Re, OS, Ir, PT, Au and Hg (Atomic No. = 80).

59.	Silver bromide is used in photography.(a) True(b) FalseI had been asked this question during a job interview.This key question is from Application of d and f-Block Elements in division D and F-Block Elements of Chemistry – Class 12
Answer» The correct answer is (a) True The explanation: Silver bromide (AgBr), a SOFT, pale-yellow, water-insoluble salt WELL KNOWN (ALONG with other silver HALIDES) for its unusual sensitivity to light. AgBr is widely used in photographic films and is believed by some to have been used for making the Shroud of Turin.

60.	Which is the most stable oxidation state of actinoids?(a) +2(b) +3(c) +4(d) +5The question was posed to me by my school principal while I was bunking the class.I would like to ask this question from D and F-Block Elements in chapter D and F-Block Elements of Chemistry – Class 12
Answer» Correct option is (b) +3 To explain I would SAY: ACTINOIDS show variable oxidation states because of the smaller energy GAP between 5f, 6d and 7s orbitals. THOUGH 3+ is the most stable oxidation state, other oxidation states are possible because of the good shielding of f-electrons.

61.	Which of the following is not a very hard metal?(a) Chromium(b) Molybdenum(c) Tungsten(d) ZincI got this question by my school principal while I was bunking the class.The query is from General Properties of the Transition Elements (D-Block) topic in section D and F-Block Elements of Chemistry – Class 12
Answer» Correct choice is (d) Zinc To explain: Greater the NUMBER of unpaired electrons, STRONGER is the bonding due to overlapping of unpaired electrons between different metal atoms. CR, MO and W have maximum unpaired d-electrons and are harder metals whereas Zn, Cd and Hg are not very HARD due to the absence of unpaired electrons.

62.	Which is the last element of lanthanides?(a) Ytterbium(b) Lutetium(c) Thulium(d) ErbiumThe question was posed to me during an interview.My question comes from D and F-Block Elements in section D and F-Block Elements of Chemistry – Class 12
Answer» Correct answer is (B) Lutetium Explanation: The atomic numbers of ERBIUM, Thulium, Ytterbium and Lutetium is 68, 69, 70 and 71 respectfully. So the last element of lanthanide series is Lutetium with ELECTRONIC configuration [Xe]4f^145d^16s^2.

63.	The lanthanides are d-block elements.(a) True(b) FalseI have been asked this question by my college professor while I was bunking the class.Query is from D and F-Block Elements in section D and F-Block Elements of Chemistry – Class 12
Answer» The correct choice is (b) False Easiest explanation: The lanthanides are ELEMENTS in which the differentiating electron ENTERS (n-2)f-orbital of anti-penultimate SHELL and are CALLED as f-block elements. These have three outer shells incomplete. Thus, lanthanides are f-block elements, not d-block.

64.	The d-electrons affect the properties of the transition elements to a great extent.(a) True(b) FalseI had been asked this question during an interview.The doubt is from General Properties of the Transition Elements (D-Block) topic in chapter D and F-Block Elements of Chemistry – Class 12
Answer» Correct choice is (a) True Easiest explanation: The TRANSITION elements DIFFER from one ANOTHER only in the number of electrons in the d-orbitals of the penultimate shell. The d-orbitals of the transition elements project to the PERIPHERY of the atom more than s and p-orbitals. THEREFORE, the d-electrons affect the properties of transition elements to a great extent.

65.	Why does Zinc not show variable valency?(a) Complete ‘d’ subshell(b) Inert pair effect(c) 4s^2 subshell(d) 4s^3 subshellThis question was addressed to me in quiz.I'd like to ask this question from Electronic Configurations of the D-Block Elements topic in section D and F-Block Elements of Chemistry – Class 12
Answer» The correct ANSWER is (a) Complete ‘d’ subshell Explanation: Both zinc and mercury have completely FILLED d-orbital. So, they tend to lose only 2 electrons from their outermost shell and have only one OXIDATION state. HENCE, they only show a valency of +2 and do not show VARIABLE valency.

66.	How many series’ of transition elements are present in the periodic table?(a) One(b) Two(c) Three(d) FourThe question was posed to me in an international level competition.I would like to ask this question from Position of the Periodic Table topic in section D and F-Block Elements of Chemistry – Class 12
Answer» CORRECT CHOICE is (d) Four Best explanation: Unlike s or p block ELEMENTS which are USUALLY discussed as columns or groups, d-block elements are better discussed by classifying them into horizontal series. In the periodic table, there are four main TRANSITION series of elements corresponding to filling of 3d, 4d, 5d and 6d sublevels in the 4^th, 5^th, 6^th and 7^th periods.

67.	All d-block elements are transition elements.(a) True(b) FalseThe question was asked during an interview.The origin of the question is Position of the Periodic Table topic in section D and F-Block Elements of Chemistry – Class 12
Answer» Right answer is (B) False The explanation: Scandium and ZINC are d-block elements but are not TRANSITION elements. The electronic configurations of Sc and ZN are: Sc: [Ar] 3d^14s^2 Zn: [Ar] 3d^104s^2 In their IONIC form, Sc and Zn lose electrons and attain the configuration of Argon. Hence, they have completely filled d orbitals.

68.	Sc^3+ is a paramagnetic ion.(a) True(b) FalseI have been asked this question during an interview.Asked question is from D and F-Block Elements topic in section D and F-Block Elements of Chemistry – Class 12
Answer» Correct answer is (b) False Easy explanation: Scandium (ATOMIC no.21, symbol-Sc) has the FOLLOWING electronic CONFIGURATION: [Ar] 3d^14s^2 In its Sc^3+ state, scandium LOSES the 4S and 3d electrons and is colourless. Hence, Sc^3+ is diamagnetic and not paramagnetic.

69.	Why is there an increase in the atomic radius of transition elements at the end of the period?(a) Increased electron-electron repulsions(b) Decreased electron-electron repulsions(c) Increase in nuclear charge(d) Increase in atomic massThis question was posed to me during an online interview.My question is taken from General Properties of the Transition Elements (D-Block) topic in division D and F-Block Elements of Chemistry – Class 12
Answer» Right CHOICE is (a) Increased electron-electron repulsions For explanation: Near the END of the period, the increased electron-electron repulsions between added ELECTRONS in the same orbitals are greater than the attractive forces due to increased nuclear CHARGE. This results in the expansion of the electron cloud and therefore, increases the atomic SIZE.