Free Atomic Structure MCQs with Answers

31. Chromium and copper have anomalous electronic configurations because

• A. their nuclei are unstable
• B. a half filled or completely filled d subshell is more stable
• C. they lose electrons easily
• D. the 4s orbital does not exist for them

Explanation: Promoting one 4s electron gives chromium a 3d5 4s1 arrangement and copper a 3d10 4s1 arrangement, both of which gain stability from the symmetry and reduced repulsion of a half filled or full subshell. These two exceptions are examined far more often than the rule itself. The energy gap between 4s and 3d is small enough for the swap to be worthwhile.

32. The energy required to remove the most loosely bound electron from a gaseous atom is called

• A. electron affinity
• B. ionisation energy
• C. electronegativity
• D. lattice energy

Explanation: Ionisation energy is always endothermic because work must be done against the nuclear attraction, and it rises steeply for each successive electron removed. Electron affinity is the energy change when an electron is added instead, and electronegativity describes the pull on shared electrons within a bond. All three are distinct and are regularly confused.

33. Which set of quantum numbers is not allowed for an electron?

• A. n=2, l=1, m=0, s=+1/2
• B. n=3, l=0, m=0, s=-1/2
• C. n=2, l=2, m=1, s=+1/2
• D. n=1, l=0, m=0, s=+1/2

Explanation: The azimuthal quantum number can only run from 0 up to n minus 1, so with n equal to 2 the value of l cannot reach 2, and the set describes a non existent 2d orbital. The other three sets satisfy every restriction, including that m lies between minus l and plus l. Checking l against n first, then m against l, resolves questions of this type quickly.

34. Shape of orbital is determined by quantum number:

• A. n
• B. l
• C. m
• D. s

Explanation: The azimuthal quantum number l fixes the subshell and therefore the shape, with l equal to 0 giving a spherical s orbital and l equal to 1 giving a dumb-bell shaped p orbital. The principal quantum number n sets the main energy level and size, m sets the orientation in space and s the direction of electron spin. Knowing which quantum number controls which property answers a whole family of questions.

35. The value of Planck's constant is:

• A. 6.6262x10-30 J.sec
• B. 6.6262x10-32 J.sec
• C. 6.6262x10+34 J.sec
• D. 6.6262x10-34 J.sec

Explanation: Planck's constant is 6.626 times 10 to the minus 34 joule seconds, and its extreme smallness is why quantisation is invisible in everyday life and only matters at atomic scale. The positive exponent in one option would make it enormous, which a moment's thought rules out. It appears in E equals hf, in the de Broglie relation and in the uncertainty principle.

36. Number of unpaired electrons present in the ground state of Fe3+ are (Atomic number of Fe = 26).

• A. Three
• B. Four
• C. Five
• D. Six

Explanation: Neutral iron is [Ar] 3d6 4s2, and forming Fe3+ removes both 4s electrons and one 3d electron, leaving 3d5. By Hund's rule those five electrons occupy the five d orbitals singly with parallel spins, so all five are unpaired, and that half filled arrangement is what makes Fe3+ so stable. Removing d electrons before the 4s pair is the standard error.

37. Number of unpaired electrons in boron in ground state is/are:

• A. 1
• B. 2
• C. 3
• D. 4

Explanation: Boron has atomic number 5, so its configuration is 1s2 2s2 2p1 and the single 2p electron is the only unpaired one. The 1s and 2s electrons are paired within their orbitals. Boron's three valence electrons allow three bonds, but only one of them is unpaired in the ground state, which is the distinction being tested.