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Atomic Struc

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卡片总数: 24内容版本: v4公开卡包更新时间: 8/1/2026

卡片预览 (24 张)

#1
正面 (问题)

Defn of atom

背面 (解答)

Atom is the smallest particle found in an element that can take part in a chemical rctn

#2
正面 (问题)

Subatomic particle - location, relative mass, relative charge, symbol

背面 (解答)

E- in orbitals around nucleus, 1/1840 , -1, (0 -1) e Neutron - nucleus, 1, 0, (1 0) n Proton -nuclues, 1, +1, (1 1) p

#3
正面 (问题)

Behaviour of subatomic particles in an electric field (3)

背面 (解答)

• Same speed • diff direction of deflection • diff angle of deflection

#4
正面 (问题)

how to find angle of deflection and stuff (formula)

背面 (解答)

angle of deflection prop. to | q/m | when solving qn use the constant k

#5
正面 (问题)

what is a nuclide

背面 (解答)

any species of given mass no. and atomic no. (eg hydrogen - 1) - elemental name & mass no

#6
正面 (问题)

What is nucleon no.

背面 (解答)

mass no

#7
正面 (问题)

what is mass no.

背面 (解答)

nucleon no.

#8
正面 (问题)

What is atomic no.

背面 (解答)

proton no.

#9
正面 (问题)

sig of atomic no.

背面 (解答)

determines the identity of an atom

#10
正面 (问题)

isotopes & their properties

背面 (解答)

same no. of protons/e- but diff neutrons - same chem properties n diff/masses/phy properties

#11
正面 (问题)

How to solve (235 92) U + (1 0) n -> Y + (90 36) Kr + 2 (1 0) n nucleon no. of Y = Proton no. of Y = Hence Y is =

背面 (解答)

Nucleon no. of Y = 235+1-90-2 = 144 Proton no. of Y = 92-36 = 56 Hence Y = (144 56) Ba

#12
正面 (问题)

Principle quantum shell, All the subshells, no. of orbitals, types of orbitals - name, shape, direction, size

背面 (解答)

Subshells: s, p, d, f S: 1 - spherical shape, non directional P: 3 -dumbbell shape, directional as the e- density is concentrated in certain directed along…axes D: 5 - d xz & d xy & d yz - 3 orbitals with similar 4-loped shape, orbitals have their lobes pointing b/w the axes d x^2 - y^2 - 4 lobed shape, lobes aligned along the x & y axes d z^2 - dumbbell shaped surrounded by a small doughnut shaped ring at its waist, orbital is aligned along the z axis

#13
正面 (问题)

the greater the value of n … (4)

背面 (解答)

further the shell is from the nucleus, higher the energy lvl of the shell higher the energy lvl of the shell/e- weaker the electrostatic attraction b/w nuceus and e- larger size of orbital

#14
正面 (问题)

energy lvl diagram & the exception

背面 (解答)

4s<3d

#15
正面 (问题)

3 basic rules to write e- configuration

背面 (解答)

Aufbau principle - e- fill orbitals from the lowest energy orbital upwards Hund’s rule - orbitals of a subshell must be occupied singly by an e- of parallel spins before pairing can occur Pauli exclusion principle - each orbital can hold a max of 2 e- and they must be of opp. spins

#16
正面 (问题)

why are paired e- stable

背面 (解答)

when they spin in opp directions, the magnetic attraction which results from their opp spins can counterbalance the electrical repulsion which results from their identical charges

#17
正面 (问题)

2 anomalous e- configuration and why are they more stable ( general reason and indiv reason)

背面 (解答)

Cr & Cu - a ‘d’ subshell that is half-fulled or full is more stable - for Cr: as 3d & 4s are abt equal in energy by the time Cr is reached, so by having 1 e- each in the 4d and 4s orbitals, inter-electronic repulsion is minimized For Cu: the fully filled 3d subshell is unusually stable due to the symmetrical charge distribution around the metal center

#18
正面 (问题)

excited state

背面 (解答)

one or more e- absorb energy and are promoted to a higher energy lvl

#19
正面 (问题)

are 4s e- lost before or after 3d e-? and why

背面 (解答)

once e- occupy the inner 3d orbitals, they provide some shielding for the outermost 4s e- hence they repel the 4s e- to a slightly higher energy lvl

#20
正面 (问题)

isoelectronic species

背面 (解答)

species with the same total no. of e-

#21
正面 (问题)

e- configuration from the periodic table

背面 (解答)

s- block - ns^1 ns&2 d-block - (n-1)d1 ns^y - the 1 is starting from the first transition metal p-block - ns^2np^1 -> ns^2np^6

#22
正面 (问题)

Defn of atomic radius

背面 (解答)

Half the shortest inter-nuclear dist found in the struc of the element

#23
正面 (问题)

Variation in atomic radii across period

背面 (解答)

DECREASE as No.of e- shells remain the same no. of protons, hence nuclear charge increases e- increase but since they are added to the valence shell, the shielding effect remains approx const effective nuclear charge increases electrostatic attraction b/w nucleus and valence e- increases decrease in size of e- cloud

#24
正面 (问题)

Variation in atomic radii down grp

背面 (解答)

INCREASE although protons increase, nuclear charge increases no. of e- shell increases, dist b/w nucleus and valence e- increase hence, the electrostatic attraction b/w the nucleus and the valence e- decreased increase in size of e- cloud