Atomic Structure — Practice Quiz

A completely filled 3d subshell (3d¹⁰) provides extra stability due to symmetrical distribution of electrons and higher exchange energy.

Step 1: Aufbau principle predicts [Ar] 3d⁹ 4s² for Cu, but the observed configuration is [Ar] 3d¹⁰ 4s¹. Step 2: A completely filled subshell (3d¹⁰) is highly symmetrical. Symmetry in electron distribution lowers the overall energy of the atom, making it more stable. Step 3: Exchange energy — in 3d¹⁰ all orbitals are filled, and the number of possible exchanges between parallel-spin electrons in 3d¹⁰ is greater than in 3d⁹, leading to higher exchange energy stabilisation. Step 4: Option A is wrong — energy ordering of 4s and 3d depends on the element and electron-electron repulsions. Option C i

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Step 1: The actual (observed) electronic configuration of Cr (Z=24) is [Ar] 3d⁵ 4s¹ — NOT the Aufbau-predicted [Ar] 3d⁴ 4s² — due to extra stability of half-filled 3d subshell. Step 2: In [Ar] 3d⁵ 4s¹: the 3d subshell has 5 electrons, one in each of the five 3d orbitals (by Hund's rule), all with the same spin — so 5 unpaired electrons. The 4s subshell has 1 electron — also unpaired. Step 3: Total unpaired electrons = 5 (from 3d) + 1 (from 4s) = 6. Step 4: If we mistakenly used [Ar] 3d⁴ 4s², we would get 4 unpaired (3d⁴) + 0 unpaired (4s² paired) = 4 — which is option A, a common wrong answer

The nucleus occupies a very small volume relative to the atom but contains most of the atom's mass and all of its positive charge.

Step 1: Most alpha particles (positively charged He²⁺ ions) passed straight through the gold foil without deflection. This means most of the atom is empty space. Step 2: A very small fraction (~1 in 10,000) bounced back. This means they encountered an extremely dense, small region of concentrated positive charge — the nucleus. Step 3: The rarity of large deflections tells us the nucleus is extremely small relative to the atom's size (nucleus ~10⁻¹⁵ m vs atom ~10⁻¹⁰ m). Step 4: Option A is wrong — Thomson's model proposed spread-out positive charge, which Rutherford's experiment disproved. Opti

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Step 1: The formula for the number of radial (spherical) nodes in an orbital is: Radial nodes = n − l − 1. Step 2: For a 4p orbital: n = 4, l = 1 (since p corresponds to l = 1). Step 3: Radial nodes = 4 − 1 − 1 = 2. Step 4: Additionally, the total number of nodes = n − 1 = 3. Out of these, angular (nodal plane) nodes = l = 1. So radial nodes = total nodes − angular nodes = 3 − 1 = 2. This confirms our answer. Common error: students confuse radial nodes with total nodes (n−1 = 3) or angular nodes (l = 1).